- Thread starter
- #1

- Thread starter shorty888
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- Thread starter
- #1

- Jan 26, 2012

- 890

Since \(1/8=2^{-3}\), you want to solve:Solve 2^(4x-3)=1/8, find the exact answer

\(2^{4x-3}=2^{-3}\)

so, now can you do it?

CB

- Feb 13, 2012

- 1,704

Compute logarithm base 2 of both terms...Solve 2^(4x-3)=1/8, find the exact answer

Kind regards

$\chi$ $\sigma$

- Thread starter
- #4

- Feb 13, 2012

- 1,704

You have $\displaystyle a=b \implies \log_{2} a= \log_{2} b$, so that $\displaystyle a=2^{4x-3},\ b=\frac{1}{8} \rightarrow 4x-3=-3$ and we have a first order algebraic equation...No, I don't understand.. I can't do it.. How??

Kind regards

$\chi$ $\sigma$

- Jan 26, 2012

- 890

In future please quote which post you are responding to (use the reply with quote option).No, I don't understand.. I can't do it.. How??

If you are refering to my post, the exponents on both sides are equal so: \(4x-3=-3\)

This is essentially the same as chisigma's method.

CB