College Algebra & Trigonometry, 2018a

3.3. SOLVING EXPONENTIAL EQUATIONS 169
If we collect like terms, we’ll end up with:
x ∗ 3log4=x ∗ 2log9− log 9
log 9 = x ∗ 2log9− x ∗ 3log4
At this point, if we want to get the x by itself, we need to factor out the x on the
right-hand side:
log 9 = x ∗ 2log9− x ∗ 3log4
log 9 = x(2 log 9 − 3 log 4)
Then divide on both sides by the coefficient in parentheses:
log 9
2log9− 3log4 = x ✭✭✭✭✭✭✭✭✭
(2 log 9 − 3 log 4)
✭✭✭✭✭✭✭✭
2log9− 3log4
log 9
2log9− 3log4 = x
9.327 ≈ x
Again, we can check our answer by plugging it back into the equation:
4 3∗9.327 ≈ 4 27.981 ≈ 7.0184 ∗ 10 16
9 2∗9.327−1 ≈ 9 17.654 ≈ 7.0177 ∗ 10 16