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 33 Page 339

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_2 x + log_2(x-2) = 3

log_2(m) + log_2(n)=log_2(mn)

log_2 (x(x-2)) = 3

m=log_2(2^m)

log_2 (x(x-2)) = log_2 2^3
log_2 (x(x-2)) = log_2 8
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_2 (x(x-2)) = log_2 8

Equate arguments

x(x-2) = 8
â–Ľ
Solve for x
x^2-2x = 8
x^2-2x-8=0
x^2+2x-4x-8=0
x(x+2)-4x-8=0
x(x+2)-4(x+2)=0
(x+2)(x-4)=0
lcx+2=0 & (I) x-4=0 & (II)
lx=-2 x-4=0
lx=-2 x=4
To check for extraneous solutions, we will substitute both -2 and 4 for x in the given equation one at a time.
log_2 x + log_2(x-2) = 3
log_2 ( -2) + log_2( -2-2) ? = 3
log_2 (-2) + log_2(-4) ? = 3 *
The argument of a logarithmic function cannot be negative, so both log_2(-2) and log_2(-4) are undefined, and -4 is an extraneous solution. Let's now check for x=4.
log_2 x + log_2(x-2) = 3
log_2 4 + log_2( 4-2) ? = 3
log_2 4 + log_2 2 ? = 3

log_2(2) = 1

log_2 4 + 1 ? = 3

Calculate logarithm

2 + 1 ? = 3
3 = 3 âś“
Since substituting 4 for x in the given equation produces a true statement, x=4 is the solution to our equation.