omitting all the base 8 stuff, we have
1/3 log(x+1) = 2log3 - 2/3 log(x+1)
log(x+1) = 2log3
log(x+1) = log9
not sure what the extra = means.
(2^x + 8*2^x)/2 = 3
9*2^x/2 = 3
9*2^x = 6
2^x = 3/2
x = ln(3/2)/ln2
May be a typo here; it's an odd answer
omitting all the base a stuff,
log(18x^3) = log18 + loga^3
= 2log3 + log2 + 3loga
oops it was 2^-x
im confused why would you omit the bases?
just for readability. Got tired of typing it in. And, until you need to resolve the base, it doesn't really matter. In #1, the equations work regardless of base, since we are working with logs on both sides of the equation.
In #3, the base didn't matter until we could use it to say that log_a(a) = 1
For #2, I suspected it might have been 2^-x, and it helps a lot!
(2^x + 8*2^-x)/2 = 3
2^x + 8/2^x = 6
if you let u = 2^x, then you have
u+8/u = 6
u^2 + 8 = 6u
u^2 - 6u + 8 = 0
(u-4)(u-2) = 0
u=4 or u=2
2^x=4 or 2^x=2
x=2 or x=1
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