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3.2 more on log and exponential equations 3.2 more on log and exponential equations Presentation Transcript

  • More on Log and Exponential Equations
  • More on Log and Exponential EquationsIn this section we continue with log-equations andexp-equations.
  • More on Log and Exponential EquationsIn this section we continue with log-equations andexp-equations. They may be grouped in two types. View slide
  • More on Log and Exponential EquationsIn this section we continue with log-equations andexp-equations. They may be grouped in two types.* Equations that do not need calculators (Equations with related bases on both sides). View slide
  • More on Log and Exponential EquationsIn this section we continue with log-equations andexp-equations. They may be grouped in two types.* Equations that do not need calculators (Equations with related bases on both sides).* Numerical equations that require calculators.
  • More on Log and Exponential EquationsIn this section we continue with log-equations andexp-equations. They may be grouped in two types.* Equations that do not need calculators (Equations with related bases on both sides).* Numerical equations that require calculators. Equations That Do Not Need Calculators
  • More on Log and Exponential EquationsIn this section we continue with log-equations andexp-equations. They may be grouped in two types.* Equations that do not need calculators (Equations with related bases on both sides).* Numerical equations that require calculators. Equations That Do Not Need CalculatorsThe Law of Uniqueness of Log and Exp Functions:
  • More on Log and Exponential EquationsIn this section we continue with log-equations andexp-equations. They may be grouped in two types.* Equations that do not need calculators (Equations with related bases on both sides).* Numerical equations that require calculators. Equations That Do Not Need CalculatorsThe Law of Uniqueness of Log and Exp Functions: If logb(U) = logb(V) then U = V.
  • More on Log and Exponential EquationsIn this section we continue with log-equations andexp-equations. They may be grouped in two types.* Equations that do not need calculators (Equations with related bases on both sides).* Numerical equations that require calculators. Equations That Do Not Need CalculatorsThe Law of Uniqueness of Log and Exp Functions: If logb(U) = logb(V) then U = V. If expb(U) = expb(V) then U = V.
  • More on Log and Exponential EquationsIn this section we continue with log-equations andexp-equations. They may be grouped in two types.* Equations that do not need calculators (Equations with related bases on both sides).* Numerical equations that require calculators. Equations That Do Not Need CalculatorsThe Law of Uniqueness of Log and Exp Functions: If logb(U) = logb(V) then U = V. If expb(U) = expb(V) then U = V.Use this law to simplify log or exp equations whenboth sides can be consolidated into a common base.
  • More on Log and Exponential Equations In this section we continue with log-equations and exp-equations. They may be grouped in two types. * Equations that do not need calculators (Equations with related bases on both sides). * Numerical equations that require calculators. Equations That Do Not Need Calculators The Law of Uniqueness of Log and Exp Functions: If logb(U) = logb(V) then U = V. If expb(U) = expb(V) then U = V.Use this law to simplify log or exp equations whenboth sides can be consolidated into a common base.An example of numbers with common base are{.., 1/8, ¼, ½, 1, 2, 4, 8,..}.
  • More on Log and Exponential Equations In this section we continue with log-equations and exp-equations. They may be grouped in two types. * Equations that do not need calculators (Equations with related bases on both sides). * Numerical equations that require calculators. Equations That Do Not Need Calculators The Law of Uniqueness of Log and Exp Functions: If logb(U) = logb(V) then U = V. If expb(U) = expb(V) then U = V.Use this law to simplify log or exp equations whenboth sides can be consolidated into a common base.An example of numbers with common base are{.., 1/8, ¼, ½, 1, 2, 4, 8,..}. All these numbers arebase 2 numbers, i.e. they are powers of 2.
  • Equations That Do Not Need CalculatorsFor exp-equations of this type, put all the bases into acommon base first.
  • Equations That Do Not Need CalculatorsFor exp-equations of this type, put all the bases into acommon base first. Then consolidate the exponentson each side
  • Equations That Do Not Need CalculatorsFor exp-equations of this type, put all the bases into acommon base first. Then consolidate the exponentson each side and put the equation into the formb u = bv ,
  • Equations That Do Not Need CalculatorsFor exp-equations of this type, put all the bases into acommon base first. Then consolidate the exponentson each side and put the equation into the formbu = bv, then drop the base b and solve U = V.
  • Equations That Do Not Need CalculatorsFor exp-equations of this type, put all the bases into acommon base first. Then consolidate the exponentson each side and put the equation into the formbu = bv, then drop the base b and solve U = V.Example A: Solve 2*42x – 1 = 81 – 3x
  • Equations That Do Not Need CalculatorsFor exp-equations of this type, put all the bases into acommon base first. Then consolidate the exponentson each side and put the equation into the formbu = bv, then drop the base b and solve U = V.Example A: Solve 2*42x – 1 = 81 – 3xSince 4 = 22 and 8 = 23,
  • Equations That Do Not Need CalculatorsFor exp-equations of this type, put all the bases into acommon base first. Then consolidate the exponentson each side and put the equation into the formbu = bv, then drop the base b and solve U = V.Example A: Solve 2*42x – 1 = 81 – 3xSince 4 = 22 and 8 = 23, put both sides of the equationin base 2 as
  • Equations That Do Not Need CalculatorsFor exp-equations of this type, put all the bases into acommon base first. Then consolidate the exponentson each side and put the equation into the formbu = bv, then drop the base b and solve U = V.Example A: Solve 2*42x – 1 = 81 – 3xSince 4 = 22 and 8 = 23, put both sides of the equationin base 2 as 2*42x – 1 = 81 – 3x 2*(22)2x – 1 = (23)1 – 3x
  • Equations That Do Not Need CalculatorsFor exp-equations of this type, put all the bases into acommon base first. Then consolidate the exponentson each side and put the equation into the formbu = bv, then drop the base b and solve U = V.Example A: Solve 2*42x – 1 = 81 – 3xSince 4 = 22 and 8 = 23, put both sides of the equationin base 2 as 2*42x – 1 = 81 – 3x 2*(22)2x – 1 = (23)1 – 3x 2*24x – 2 = 23 – 9x
  • Equations That Do Not Need CalculatorsFor exp-equations of this type, put all the bases into acommon base first. Then consolidate the exponentson each side and put the equation into the formbu = bv, then drop the base b and solve U = V.Example A: Solve 2*42x – 1 = 81 – 3xSince 4 = 22 and 8 = 23, put both sides of the equationin base 2 as 2*42x – 1 = 81 – 3x 2*(22)2x – 1 = (23)1 – 3x 2*24x – 2 = 23 – 9x 21 + 4x – 2 = 23 – 9x
  • Equations That Do Not Need CalculatorsFor exp-equations of this type, put all the bases into acommon base first. Then consolidate the exponentson each side and put the equation into the formbu = bv, then drop the base b and solve U = V.Example A: Solve 2*42x – 1 = 81 – 3xSince 4 = 22 and 8 = 23, put both sides of the equationin base 2 as 2*42x – 1 = 81 – 3x 2*(22)2x – 1 = (23)1 – 3x 2*24x – 2 = 23 – 9x 21 + 4x – 2 = 23 – 9x drop the base 2 1 + 4x – 2 = 3 – 9x
  • Equations That Do Not Need CalculatorsFor exp-equations of this type, put all the bases into acommon base first. Then consolidate the exponentson each side and put the equation into the formbu = bv, then drop the base b and solve U = V.Example A: Solve 2*42x – 1 = 81 – 3xSince 4 = 22 and 8 = 23, put both sides of the equationin base 2 as 2*42x – 1 = 81 – 3x 2*(22)2x – 1 = (23)1 – 3x 2*24x – 2 = 23 – 9x 21 + 4x – 2 = 23 – 9x drop the base 2 1 + 4x – 2 = 3 – 9x 4x – 1 = 3 – 9x 13x = 4  x = 4/13
  • Equations That Do Not Need CalculatorsFor log-equations of this type, consolidate the logs oneach side first.
  • Equations That Do Not Need CalculatorsFor log-equations of this type, consolidate the logs oneach side first. There are two following possibilities.
  • Equations That Do Not Need CalculatorsFor log-equations of this type, consolidate the logs oneach side first. There are two following possibilities.I. If the resulting equation is of the form logb(U) = V,
  • Equations That Do Not Need CalculatorsFor log-equations of this type, consolidate the logs oneach side first. There are two following possibilities.I. If the resulting equation is of the form logb(U) = V,drop the base b by writing it in the exponential formU = b V.
  • Equations That Do Not Need CalculatorsFor log-equations of this type, consolidate the logs oneach side first. There are two following possibilities.I. If the resulting equation is of the form logb(U) = V,drop the base b by writing it in the exponential formU = b V.Example B: Solve log2(x – 1) + log2(x + 3) = 5
  • Equations That Do Not Need CalculatorsFor log-equations of this type, consolidate the logs oneach side first. There are two following possibilities.I. If the resulting equation is of the form logb(U) = V,drop the base b by writing it in the exponential formU = b V.Example B: Solve log2(x – 1) + log2(x + 3) = 5Combine the log using product rule: log2[(x – 1)(x + 3)] = 5
  • Equations That Do Not Need CalculatorsFor log-equations of this type, consolidate the logs oneach side first. There are two following possibilities.I. If the resulting equation is of the form logb(U) = V,drop the base b by writing it in the exponential formU = b V.Example B: Solve log2(x – 1) + log2(x + 3) = 5Combine the log using product rule: log2[(x – 1)(x + 3)] = 5Write it in exp-form: (x – 1)(x + 3) = exp2(5) = 25
  • Equations That Do Not Need CalculatorsFor log-equations of this type, consolidate the logs oneach side first. There are two following possibilities.I. If the resulting equation is of the form logb(U) = V,drop the base b by writing it in the exponential formU = b V.Example B: Solve log2(x – 1) + log2(x + 3) = 5Combine the log using product rule: log2[(x – 1)(x + 3)] = 5Write it in exp-form: (x – 1)(x + 3) = exp2(5) = 25 x2 + 2x – 3 = 32
  • Equations That Do Not Need CalculatorsFor log-equations of this type, consolidate the logs oneach side first. There are two following possibilities.I. If the resulting equation is of the form logb(U) = V,drop the base b by writing it in the exponential formU = b V.Example B: Solve log2(x – 1) + log2(x + 3) = 5Combine the log using product rule: log2[(x – 1)(x + 3)] = 5Write it in exp-form: (x – 1)(x + 3) = exp2(5) = 25 x2 + 2x – 3 = 32 x2 + 2x – 35 = 0
  • Equations That Do Not Need CalculatorsFor log-equations of this type, consolidate the logs oneach side first. There are two following possibilities.I. If the resulting equation is of the form logb(U) = V,drop the base b by writing it in the exponential formU = b V.Example B: Solve log2(x – 1) + log2(x + 3) = 5Combine the log using product rule: log2[(x – 1)(x + 3)] = 5Write it in exp-form: (x – 1)(x + 3) = exp2(5) = 25 x2 + 2x – 3 = 32 x2 + 2x – 35 = 0 (x + 7)(x – 5) = 0
  • Equations That Do Not Need CalculatorsFor log-equations of this type, consolidate the logs oneach side first. There are two following possibilities.I. If the resulting equation is of the form logb(U) = V,drop the base b by writing it in the exponential formU = b V.Example B: Solve log2(x – 1) + log2(x + 3) = 5Combine the log using product rule: log2[(x – 1)(x + 3)] = 5Write it in exp-form: (x – 1)(x + 3) = exp2(5) = 25 x2 + 2x – 3 = 32 x2 + 2x – 35 = 0 (x + 7)(x – 5) = 0 x = -7, x = 5
  • Equations That Do Not Need CalculatorsFor log-equations of this type, consolidate the logs oneach side first. There are two following possibilities.I. If the resulting equation is of the form logb(U) = V,drop the base b by writing it in the exponential formU = b V.Example B: Solve log2(x – 1) + log2(x + 3) = 5Combine the log using product rule: log2[(x – 1)(x + 3)] = 5Write it in exp-form: (x – 1)(x + 3) = exp2(5) = 25 x2 + 2x – 3 = 32 x2 + 2x – 35 = 0 (x + 7)(x – 5) = 0 x = -7, x = 5
  • Equations That Do Not Need CalculatorsII. If after consolidating the logs, the resultingequation is of the form logb(U) = logb (V),
  • Equations That Do Not Need CalculatorsII. If after consolidating the logs, the resultingequation is of the form logb(U) = logb (V), just drop thelog and solve the equation U = V.
  • Equations That Do Not Need CalculatorsII. If after consolidating the logs, the resultingequation is of the form logb(U) = logb (V), just drop thelog and solve the equation U = V.Example C: Solve log4(x + 2) – log4(x – 1) = log4 (5)
  • Equations That Do Not Need CalculatorsII. If after consolidating the logs, the resultingequation is of the form logb(U) = logb (V), just drop thelog and solve the equation U = V.Example C: Solve log4(x + 2) – log4(x – 1) = log4 (5)Combine the logs using quotient rule: (x + 2) log4[ ] = log4(5) (x – 1)
  • Equations That Do Not Need CalculatorsII. If after consolidating the logs, the resultingequation is of the form logb(U) = logb (V), just drop thelog and solve the equation U = V.Example C: Solve log4(x + 2) – log4(x – 1) = log4 (5)Combine the logs using quotient rule: (x + 2) log4[ ] = log4(5) (x – 1)Drop the logs on both sides we get (x + 2) =5 (x – 1)
  • Equations That Do Not Need CalculatorsII. If after consolidating the logs, the resultingequation is of the form logb(U) = logb (V), just drop thelog and solve the equation U = V.Example C: Solve log4(x + 2) – log4(x – 1) = log4 (5)Combine the logs using quotient rule: (x + 2) log4[ ] = log4(5) (x – 1)Drop the logs on both sides we get (x + 2) =5 (x – 1) x + 2 = 5(x – 1)
  • Equations That Do Not Need CalculatorsII. If after consolidating the logs, the resultingequation is of the form logb(U) = logb (V), just drop thelog and solve the equation U = V.Example C: Solve log4(x + 2) – log4(x – 1) = log4 (5)Combine the logs using quotient rule: (x + 2) log4[ ] = log4(5) (x – 1)Drop the logs on both sides we get (x + 2) =5 (x – 1) x + 2 = 5(x – 1) x + 2 = 5x – 5
  • Equations That Do Not Need CalculatorsII. If after consolidating the logs, the resultingequation is of the form logb(U) = logb (V), just drop thelog and solve the equation U = V.Example C: Solve log4(x + 2) – log4(x – 1) = log4 (5)Combine the logs using quotient rule: (x + 2) log4[ ] = log4(5) (x – 1)Drop the logs on both sides we get (x + 2) =5 (x – 1) x + 2 = 5(x – 1) x + 2 = 5x – 5 7 = 4x
  • Numerical Equations That Require CalculatorsWith numerical equations we seek both the exactand approximate answers with calculators.
  • Numerical Equations That Require CalculatorsWith numerical equations we seek both the exactand approximate answers with calculators.We have seen some examples in the last section.
  • Numerical Equations That Require CalculatorsWith numerical equations we seek both the exactand approximate answers with calculators.We have seen some examples in the last section.In this section, we tackle the equations that containbases that are different from e and 10.
  • Numerical Equations That Require CalculatorsWith numerical equations we seek both the exactand approximate answers with calculators.We have seen some examples in the last section.In this section, we tackle the equations that containbases that are different from e and 10. We do this bychanging the calculation into base e with thefollowing formula.
  • Numerical Equations That Require CalculatorsWith numerical equations we seek both the exactand approximate answers with calculators.We have seen some examples in the last section.In this section, we tackle the equations that containbases that are different from e and 10. We do this bychanging the calculation into base e with thefollowing formula.Given x and a base b, weve blogb(x) = x,
  • Numerical Equations That Require CalculatorsWith numerical equations we seek both the exactand approximate answers with calculators.We have seen some examples in the last section.In this section, we tackle the equations that containbases that are different from e and 10. We do this bychanging the calculation into base e with thefollowing formula.Given x and a base b, weve blogb(x) = x,apply ln( ) to both sides: ln(blog b ) = ln(x) (x)
  • Numerical Equations That Require CalculatorsWith numerical equations we seek both the exactand approximate answers with calculators.We have seen some examples in the last section.In this section, we tackle the equations that containbases that are different from e and 10. We do this bychanging the calculation into base e with thefollowing formula.Given x and a base b, weve blogb(x) = x,apply ln( ) to both sides: ln(blog b ) = ln(x) (x)use the power rule logb(x)ln(b) = ln(x)
  • Numerical Equations That Require CalculatorsWith numerical equations we seek both the exactand approximate answers with calculators.We have seen some examples in the last section.In this section, we tackle the equations that containbases that are different from e and 10. We do this bychanging the calculation into base e with thefollowing formula.Given x and a base b, weve blogb(x) = x,apply ln( ) to both sides: ln(blog b ) = ln(x) (x)use the power rule logb(x)ln(b) = ln(x)Therefore we get the Change of Bases Formula ln(x) logb(x) = ln(b)
  • Numerical Equations That Require CalculatorsTo solve equations in bases different from e,
  • Numerical Equations That Require CalculatorsTo solve equations in bases different from e, weobtain the exact solution in the given base first,
  • Numerical Equations That Require CalculatorsTo solve equations in bases different from e, weobtain the exact solution in the given base first, thenuse the change of bases formula to obtain theapproximate solutions.
  • Numerical Equations That Require CalculatorsTo solve equations in bases different from e, weobtain the exact solution in the given base first, thenuse the change of bases formula to obtain theapproximate solutions.Example D: (Change of Base)Solve for x if 14 = 3·52x+1.
  • Numerical Equations That Require CalculatorsTo solve equations in bases different from e, weobtain the exact solution in the given base first, thenuse the change of bases formula to obtain theapproximate solutions.Example D: (Change of Base)Solve for x if 14 = 3·52x+1.Isolate the exp-part: 14/3 = 52x+1
  • Numerical Equations That Require CalculatorsTo solve equations in bases different from e, weobtain the exact solution in the given base first, thenuse the change of bases formula to obtain theapproximate solutions.Example D: (Change of Base)Solve for x if 14 = 3·52x+1.Isolate the exp-part: 14/3 = 52x+1 14/3 = exp5(2x+1)
  • Numerical Equations That Require CalculatorsTo solve equations in bases different from e, weobtain the exact solution in the given base first, thenuse the change of bases formula to obtain theapproximate solutions.Example D: (Change of Base)Solve for x if 14 = 3·52x+1.Isolate the exp-part: 14/3 = 52x+1 14/3 = exp5(2x+1)Rewrite in log5( ): log5(14/3) = 2x+1
  • Numerical Equations That Require CalculatorsTo solve equations in bases different from e, weobtain the exact solution in the given base first, thenuse the change of bases formula to obtain theapproximate solutions.Example D: (Change of Base)Solve for x if 14 = 3·52x+1.Isolate the exp-part: 14/3 = 52x+1 14/3 = exp5(2x+1)Rewrite in log5( ): log5(14/3) = 2x+1Solve for x: log5(14/3) – 1 = 2x
  • Numerical Equations That Require CalculatorsTo solve equations in bases different from e, weobtain the exact solution in the given base first, thenuse the change of bases formula to obtain theapproximate solutions.Example D: (Change of Base)Solve for x if 14 = 3·52x+1.Isolate the exp-part: 14/3 = 52x+1 14/3 = exp5(2x+1)Rewrite in log5( ): log5(14/3) = 2x+1Solve for x: log5(14/3) – 1 = 2x (log5(14/3) – 1) / 2 = x
  • Numerical Equations That Require CalculatorsTo solve equations in bases different from e, weobtain the exact solution in the given base first, thenuse the change of bases formula to obtain theapproximate solutions.Example D: (Change of Base)Solve for x if 14 = 3·52x+1.Isolate the exp-part: 14/3 = 52x+1 14/3 = exp5(2x+1)Rewrite in log5( ): log5(14/3) = 2x+1Solve for x: log5(14/3) – 1 = 2x (log5(14/3) – 1) / 2 = xBy base-change ln(14/3)formula using ln: ( ln(5) – 1) / 2 = x
  • Numerical Equations That Require CalculatorsTo solve equations in bases different from e, weobtain the exact solution in the given base first, thenuse the change of bases formula to obtain theapproximate solutions.Example D: (Change of Base)Solve for x if 14 = 3·52x+1.Isolate the exp-part: 14/3 = 52x+1 14/3 = exp5(2x+1)Rewrite in log5( ): log5(14/3) = 2x+1Solve for x: log5(14/3) – 1 = 2x (log5(14/3) – 1) / 2 = xBy base-change ln(14/3)formula using ln: ( ln(5) – 1) / 2 = x ≈ -0.0214
  • Numerical Equations That Require CalculatorsExample E: (Change of Base)Solve for x if 4x–1 = 72x+1.
  • Numerical Equations That Require CalculatorsExample E: (Change of Base)Solve for x if 4x–1 = 72x+1.Apply log4( ) to both sides:
  • Numerical Equations That Require CalculatorsExample E: (Change of Base)Solve for x if 4x–1 = 72x+1.Apply log4( ) to both sides: log4(4x–1) = log4(72x+1)
  • Numerical Equations That Require CalculatorsExample E: (Change of Base)Solve for x if 4x–1 = 72x+1.Apply log4( ) to both sides: log4(4x–1) = log4(72x+1) x – 1 = log4(7)(2x + 1)
  • Numerical Equations That Require CalculatorsExample E: (Change of Base)Solve for x if 4x–1 = 72x+1.Apply log4( ) to both sides: log4(4x–1) = log4(72x+1) x – 1 = log4(7)(2x + 1) x – 1 = 2log4(7)x + log4(7)
  • Numerical Equations That Require CalculatorsExample E: (Change of Base)Solve for x if 4x–1 = 72x+1.Apply log4( ) to both sides: log4(4x–1) = log4(72x+1) x – 1 = log4(7)(2x + 1) x – 1 = 2log4(7)x + log4(7)Solve for x: x – 2log4(7)x = log4(7) + 1
  • Numerical Equations That Require CalculatorsExample E: (Change of Base)Solve for x if 4x–1 = 72x+1.Apply log4( ) to both sides: log4(4x–1) = log4(72x+1) x – 1 = log4(7)(2x + 1) x – 1 = 2log4(7)x + log4(7)Solve for x: x – 2log4(7)x = log4(7) + 1 (1 – 2 log4(7))x = log4(7) + 1
  • Numerical Equations That Require CalculatorsExample E: (Change of Base)Solve for x if 4x–1 = 72x+1.Apply log4( ) to both sides: log4(4x–1) = log4(72x+1) x – 1 = log4(7)(2x + 1) x – 1 = 2log4(7)x + log4(7)Solve for x: x – 2log4(7)x = log4(7) + 1 (1 – 2 log4(7))x = log4(7) + 1 log4(7) + 1 x = 1 – 2log4(7)
  • Numerical Equations That Require CalculatorsExample E: (Change of Base)Solve for x if 4x–1 = 72x+1.Apply log4( ) to both sides: log4(4x–1) = log4(72x+1) x – 1 = log4(7)(2x + 1) x – 1 = 2log4(7)x + log4(7)Solve for x: x – 2log4(7)x = log4(7) + 1 (1 – 2 log4(7))x = log4(7) + 1 log4(7) + 1 x = 1 – 2log4(7) ln(7) ln(4) + 1Change bases x =1 – 2ln(7) ln(4)
  • Numerical Equations That Require CalculatorsExample E: (Change of Base)Solve for x if 4x–1 = 72x+1.Apply log4( ) to both sides: log4(4x–1) = log4(72x+1) x – 1 = log4(7)(2x + 1) x – 1 = 2log4(7)x + log4(7)Solve for x: x – 2log4(7)x = log4(7) + 1 (1 – 2 log4(7))x = log4(7) + 1 log4(7) + 1 x = 1 – 2log4(7) ln(7) ln(4) + 1 ln(7) + ln(4)Change bases x =1 – 2ln(7) =ln(4) – 2ln(7) ≈ -1.33 ln(4)
  • More on Log and Exponential EquationsExercise A. Solve the following equations withoutcalculators. 2. 59x– 1 = 53 21. 4 2x – 1 =4 1 – 3x 53 4. 59x– 1 = 12x 23. 4*4 2x – 1 =44 2 – 3x 55. 2*4 2x – 1 =4 1 – 3x 6. 92x+ 2(3x– 5) = 97. 2*4 = 8 x 3x 8. (2*4)x = 83x x 9 99. ( 3 )x = 33x 10. 3 = 33x
  • More on Log and Exponential EquationsExercise B. Solve. Give both the exact and theapprox. answers.11. 3x = 25 12. 7 = 4–2x+1 13. 33x–2 = 1114. 4*3x–4 = 11 15. 5x = 3–3x+1 16. 23x+1 = 32x+1Exercise C. Solve the following equations withoutcalculators.17. 2 = log2(x–1) 18. log22 = log2(x – 1)19. log(2)+log(2x–1)=1 20. log(2)–log(2x–1)=121. log3(x)+log3(x–2)=1 22. log3(x)–log3(x–2)=123. log3(x+2)+log3(x–2)=log3(5)24. log (x+2)–log (x–2)=log (5)