# Solve exponential equation.

#### shorty888

##### New member
Solve 2^(4x-3)=1/8, find the exact answer

#### CaptainBlack

##### Well-known member
Re: Solve

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

$$2^{4x-3}=2^{-3}$$

so, now can you do it?

CB

#### chisigma

##### Well-known member
Re: Solve

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

Kind regards

$\chi$ $\sigma$

#### shorty888

##### New member
Re: Solve

No, I don't understand.. I can't do it.. How??

#### chisigma

##### Well-known member
Re: Solve

No, I don't understand.. I can't do it.. How??
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...

Kind regards

$\chi$ $\sigma$

#### CaptainBlack

##### Well-known member
Re: Solve

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

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