Forgive me for this pre-calc question.

Pre-College Topics High School Life
1

<p>ill admit it, i suck at math.</p>

<p>how would you go about (step by step) solving this for x:</p>

<p>4^3x-4 = (1/8)^x-1 </p>

<p>i can only do<br>
3x-4 (log 4) = x-1 (log 1/8) , and then maybe (???)
3x-4 (log 2^2) = x-1 (log 2^-3)</p>

<p>but then i dont know what to do</p>

<p>im also stuck on 81^x+2 = 27^5x+4<br>
but if the same principles apply, there is no need to explain that one.</p>

<p>i could solve both fine if i turned it into decimals, but the book im using gives answers in fractions, so i know im not supposed to do that.</p>

<p>any help would be appreciated!!!!!!!!!!!!</p>

2

<p>is that 4^(3x-4) = (1/8)^(x-1)
?</p>

3

<p>you're on track for the first problem. You did the log correctly and brought the exponent down correctly.</p>

<p>3x-4 (log 2^2) = x-1 (log 2^-3)</p>

<p>Now notice that there are some more exponents. Bring them down like you did before to get</p>

<p>2<em>(3x-4) * log 2 = (-3)</em>(x-1) * log 2</p>

<p>Since log 2 is on both sides, you can go ahead and get rid of it by dividing by log 2. You end up with 2 * (3x-4) = (-3) * (x-1), which I'm sure you can solve the rest on your own.</p>

<p>The next problem uses the same principle</p>