Section 5.6 logarithmic and exponential equations

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Section 5.6
Logarithmic and
Exponential Equations

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Recall:

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3 3Solve: log 4 2log x=
Reminder: The domain of logarithmic functions is positive
numbers only so check for extraneous solutions.
2
3 3log 4 log x=
2
4 x=
2 or 2x x= = −

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( ) ( )2 2Solve: log 2 log 1 1x x+ + − =
( ) ( )2log 2 1 1x x+ − =
( ) ( )2 2 1x x= + −
2
2 2x x= − − +
2
0x x+ =
( )1 0x x + =
0 or 1x x= = −
Both solutions are in the
domain of the log functions.

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Recall:

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Solve: 3 7x
=
ln3 ln 7x
=
ln3 ln 7x =
ln 7
ln3
x = 1.771≈

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Solve: 5 2 3x
× =
3
2
5
x
=
3
ln 2 ln
5
x  
=  ÷
 
3
ln 2 ln
5
x
 
=  ÷
 
3
ln
5
0.737
ln 2
x
 
 ÷
 = ≈ −

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1 2 3
Solve: 2 5x x− +
=
1 2 3
ln 2 ln5x x− +
=
( ) ( )1 ln 2 2 3 ln5x x− = +
ln 2 ln 2 2 ln5 3ln5x x− = +
ln 2 2 ln5 3ln5 ln 2x x− = +
( )ln 2 2ln5 3ln5 ln 2x − = +
3ln5 ln 2
2.186
ln 2 2ln5
x
+
= ≈ −
−

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Solve: 9 3 6 0x x
− − =
( )
2
Note that 9 = 3 so the equation is quadratic in form.x x
2
3 and = 9x x
u u=
2
6 0u u− − =
( ) ( )3 2 0u u− + =
3 or 2u u= = −
3 3 or 3 2x x
= = −
1x =
3 0 for allx
x>