7.6 Solve Exponential and Logarithmic Equations

VOCABULARY

Exponential equation

An equation in which variable expressions occur as exponents

Logarithmic equation

An equation that involves logarithms of variable expressions

PROPERTY OF EQUALITY FOR EXPONENTIAL EQUATIONS

Algebra If b is a positive number other than 1, then b^x = b^y if and only if _x = y_ .

Example If 5^x = 5⁴, then x = 4. If x = _4_, then 5^x = 5⁴.

Example 1

Solve by equating exponents

Solve 64^x = 16^{x + 1}.

64^x = 16^x^+ 1	Write original equation.
(_4^3_)^x = (4^2 )^x +1	Rewrite each power with base _4_.
_4^3x= _4^2x + 2_	Power of a power property
_3x_ = _2x + 2_	Property of equality
x = _2_	Solve for x.
The solution is 2.
CHECK Substitute the solution into the original equation.
64^_2_≟ 16^_2_ ^+ 1	Substitute for x.
_4096_ = _4096_	Solution checks.

Example 2

Take a logarithm of each side

Solve 6^x = 27.

6^x = 27	Write original equation
__log6 6^x __ = _log6 27_	Take log6 of each side.
x = _log6 27	logb b^x = x
x =	Use change-of-base formula.
» _1.84_	Use a calculator.
The solution is about _1.84_. Check this in the original equation.

Example 3

Take a logarithm of each side

Solve 6e^0.25x + 8 = 20.

6e^0.25x + 8 = 20	Write original equation.
6e^0.25x = _12_	Subtract _8_ from each side.
_e^0.25x = _2_	Divide each side by _6_.
In e^0.25x = _In 2_	Take natural log of each side.
_0.25x_ = _In 2_	In e^x = loge e^x = x
x » 2.77	Divide each side by _0.25_.
The solution is about _2.77_. Check this in the original equation.

Solve the equation.

1. 3^{7x - 3} = 9^2x

1

2. 5^x = 72

2.657

3. 8^{3x + 2} - 6 = 5

20.2823

4. 3e^0.5x + 2 = 5

0

PROPERTY OF EQUALITY FOR LOGARITHMIC EQUATIONS

Algebra If b, x, and y are positive numbers with b ¹ 1,

then log_b x = log_b y if and only if _x = y .

Example If log₃ x = log₃ 8, then x = 8. If x = _8_, then log₃ x = log₃ 8.

Example 4

Solve a logarithmic equation

Solve log7(6x - 16) = log7(x - 1).
log7(6x - 16) = log7(x - 1)	Write original equation.
6x - 16 = _x - 1_	Property of equality
5x - 16 = _-1_	Subtract _x_ from each side.
_5x_ = _15_	Add _16_ to each side.
_x_ = _3_	Divide each side by _5_.
The solution is _3_.
CHECK Substitute the solution into the original equation.
log7(6x - 16) = log7(x - 1)	Write original equation.
log7( 6 · 3 -16) ≟ log7(3 - 1)	Substitute for x.
log7 2 = log7 2	Solution checks.

Example 5

Solve log₅(3x - 8) = 2.

log5(3x - 8) = 2	Write original equation.
5log5^(3x^-^8) = _5^2_	Exponentiate each side using base _5_.
3x - 8 = 25	b^logb^x = x
_3x_ = _33_	Add _8_ to each side.
_x_ = _11_	Divide each side by _3_.
The solution is _11_.
CHECK log5(3x - 8) = log5(_3 · 11 -8_) = log5 _25_ .Because _5^2_ = _25_, log5 _25_ = 2.

Example 6

Check for extraneous solutions

Solve log 5x + log(x - 1) = 2.

log 5x + log(x - 1) = 2	Write original equation.
log [ 5x (x - 1) ] = 2	Product property of logarithms
__10 ^log [5x(x^-^1)]__ = __10^2__	Exponentiate each side.
__5x(x - 1)x__ = __100__	blogb x = x
__5x^2 - 5x__ = _100_	Distributive property
__5x^2 - x - 100__ = _0_	Write in standard form.
__x^2 - x - 20__ = _0_	Divide each side by _5_.
__(x - 5) (x + 4)__ = _0_	Factor.
__x = 5_ or __x = -4__	Zero product property

CHECK x = 5 log 5 _5_ + log (_5_ - 1) ≟ 2

log _25_ + log _4_ ≟ 2

log _100_ ≟ 2

_2_ = 2

So, _5_ is a solution.

CHECK x = -4 log [5(-4)] + log (_-4_ -1) ≟ 2

log (_-20_) + log (_-5_) ≟ 2

Because log (_-20_ ) and log (_-5_)

are not defined, _-4_ is not a solution.

Solve the equation. Check for extraneous solutions.

5. ln (7x - 13)= ln (2x + 17)

6

6. log₃(2x + 9) = 3

9

7. log₄(10x + 624) = 5

40

8. log ₆(x - 9) + log₆ x= 2

12