Posted by **Rebekah** on Wednesday, November 28, 2012 at 4:30pm.

1. 1/3log base 8 of (x+1)=2log base 8 of 3-(2/3)log base 8 of (x+1)

2. 2^x+8 times 2^=x all over 2 = 3

3. if log base a of 3= x and log base a of 2 = y, find each of thefollowing in terms of x and y

log base a (18a^3)

thanks!!

- Precalc -
**Steve**, Wednesday, November 28, 2012 at 4:35pm
1.

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

x+1=9

x=8

2.

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

3.

omitting all the base a stuff,

log(18x^3) = log18 + loga^3

= log9+log2+3loga

= 2log3 + log2 + 3loga

= 2x+y+3

- Precalc -
**Rebekah**, Wednesday, November 28, 2012 at 4:48pm
oops it was 2^-x

im confused why would you omit the bases?

- Precalc -
**Steve**, Wednesday, November 28, 2012 at 4:53pm
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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