# Hi! could you please solve for x this equation? `(log8+logx)/log32=log2/(log8-logx)` Thanks

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### 3 Answers

Hi, thezy,

For this one, we could cross multiply first, giving:

(log 8 + log x)(log 8 - log x) = log 32 log 2

This simplifies to

(log 8)^2 - (log x)^2 = log 32 log 2

Then, we could subtract log 32 log 2 from each side, and add (log x)^2 to each side, giving us:

(log 8)^2 + log 32 log 2 = (log x)^2

Taking the square root of each side:

[ (log 8)^2 + log 32 log 2 ] = log x

Now, from here, we can write the log 8 = log 2^3 and log 32 = log 2^5. So:

[ (log 2^3)^2 + (log 2^5) log 2 ] = log x

Simplifying:

[ (3 log 2)^2 + (5 log 2) log 2 ] = log x

9(log 2)^2 + 5(log 2)^2 = log x

14 (log 2)^2 = log x

Taking the antilog of each side:

10^[14 (log 2)^2] = x

The left side may be able to be simplified more. I can't figure it out anymore than that, though. Sorry.

Good luck, Thezy. I hope this helps.

Till Then,

Steve

### Hide Replies ▲

Opps. Sorry. But, subtract 5(log2)log2 would give us 9-5 which if 4. So, we would have 4 instead of 14 in my final answer.

II hope that is understandable, Thezy. Sorry.

Till Then,

Steve

`rArr 9(log2)^2-(logx)^2=5(log2)^2 `

**Therefore, x=-4, 4.**

### Hide Replies ▲

What is `log(-4)?` It is not defined.

We have

**Thus we have**`x=4,1/4`

**Therefore**

**both values of x are correct.**