Big Ideas Math Algebra 2, 2014
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Big Ideas Math Algebra 2, 2014 View details
6. Solving Exponential and Logarithmic Equations
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Exercise 8 Page 336

If two equivalent logarithmic expressions have the same base, then the arguments must be equal.

x=4

Practice makes perfect
We want to solve an equation involving more than one logarithm. To do so, we will use the Product Property of Logarithms. log_b mn = log_b m + log_b n First, we will isolate the logarithm that contains the variable. Then, we can use the above property to isolate the variable from the logarithm.
log_4(x+12) + log_4x = 3

log_4(m) + log_4(n)=log_4(mn)

log_4((x+12)x) = 3
log_4(x^2+12x) = 3

m=log_4(4^m)

log_4(x^2+12x) = log_44^3
log_4(x^2+12x) = log_464
Next, we will use the fact that if two equivalent logarithmic expressions have the same base, then the arguments must be equal. log_b x=log_b y ⇔ x= y Let's apply the above property to our equation and then solve for x by factoring.
log_4(x^2+12x) = log_464

Equate arguments

x^2+12x = 64
â–Ľ
Solve for x
x^2+12x-64 = 0
x^2+16x-4x-64 = 0
x(x+16)-4(x+16) = 0
(x+16)(x-4) = 0
lcx+16=0 & (I) x-4=0 & (II)
lx=-16 x-4=0
lx=-16 x=4
To check for extraneous solutions, we will substitute both -16 and 4 for x in the given equation one at a time.
log_4(x+12) + log_4x = 3
log_4( -16+12) + log_4( -16) ? = 3
log_4(-4) + log_4(-16) ? = 3 *
The argument of a logarithmic function cannot be negative, so both log_4(-4) and log_4(-16) are undefined, and -16 is an extraneous solution. Let's now check for the x=4.
log_4(x+12) + log_4x = 3
log_4( 4+12) + log_4 4 ? = 3
log_4 16 + log_4 4 ? = 3

log_4(m) + log_4(n)=log_4(mn)

log_4 64 ? = 3

Calculate logarithm

3 = 3 âś“
Since substituting 4 for x in the given equation produces a true statement, x=4 is a solution to our equation.