1. ## Difficult logarithm question

I've been trying to do this question for 2 days! I need help. Here's the question:

8logn = logx^2/3 + log2 - 1/2logx + logx^3 - log2/x^4

2. This could be just a tip ... couldn't you simplify the logs as a first step?

3. I still don't understand.. the questions we usually get are like

log n = 1/3(log3-log5) + 2/3 log 4 + log 11

Edit: I rewrote the question like this:

logn^8 = 2/3logx +log2 - 1/2logx + 3logx - [log2 - (4logx)]

Would this help whatsoever?

4. 8logn = logx^2/3 + log2 - 1/2logx + logx^3 - log2/x^4[/quote]
8logn = logx2/3 + log2 - 1/2logx + logx3 - log2/x4
logn8 =log(x2/3 *2)-log(x1/2 *x3)-log2/x4
logn8=log2x2/3-logx7/2-log2/x4
logn8= log (2x2/3/ x7/2)- log2/x4
logn8= log2x-17/6- log2/x4
logn8= log (2x-17/6/2/x4)
logn8= log (x7/4)
n8=x9/4
n=(x7/4)1/8

5. Hello, AwesomeAndrew!

Solve for $\displaystyle n\!:\;\;8\ln n \:=\: \ln\left(x^{\frac{2}{3}}\right) + \ln(2) - \tfrac{1}{2}\ln(x) + \ln(x^3) - \ln\left(\frac{2}{x^4}\right)$

We have: .$\displaystyle 8\ln(n) \;=\;\tfrac{2}{3}\ln(x) + \ln(2) - \tfrac{1}{2}\ln(x) + 3\ln(x) - \bigg[\ln(2) - \ln(x^4)\bigg]$

. . . . . . . . . . . . $\displaystyle =\;\tfrac{2}{3}\ln(x) + {\color{red}\rlap{/////}}\ln(2) - \tfrac{1}{2}\ln(x) + 3\ln(x) - {\color{red}\rlap{/////}}\ln(2) + \ln(x^4)$

. . . . . . . . . . . . $\displaystyle =\;\tfrac{2}{3}\ln(x) - \tfrac{1}{2}\ln(x) + 3\ln(x) + 4\ln(x)$

. . . . . . . . . . . . $\displaystyle =\;\left(\tfrac{2}{3} - \tfrac{1}{2} + 3 + 4\right)\ln(x)$

. . . . . . . .$\displaystyle 8\ln(n)\;=\;\tfrac{43}{6}\ln(x)$

Hence: . . . $\displaystyle \ln(n) \:=\:\tfrac{43}{48}\ln(x) \;=\;\ln\left(x^{\frac{43}{48}}\right)$

Therefore: . . . $\displaystyle n \;=\;x^{\frac{43}{48}}$