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
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!!
3 answers
oops it was 2^-x
im confused why would you omit the bases?
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
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
1 answer