How Do You Solve the Exponential Equation 2^(4x-3) = 1/8?

[shorty888]

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

 

[CaptainBlack]

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]

Re: Solve

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

Compute logarithm base 2 of both terms...

Kind regards

$\chi$ $\sigma$

 

[shorty888]

Re: Solve

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

 

[chisigma]

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]

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

This is essentially the same as chisigma's method.

CB