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

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To solve the exponential equation 2^(4x-3) = 1/8, recognize that 1/8 can be expressed as 2^(-3). This leads to the equation 2^(4x-3) = 2^(-3), allowing for the comparison of exponents. Setting the exponents equal gives the equation 4x - 3 = -3. Solving this results in the value of x. The discussion emphasizes using logarithms and exponent properties to simplify the problem.
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Solve 2^(4x-3)=1/8, find the exact answer
 
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Re: Solve

shorty888 said:
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
 
Re: Solve

shorty888 said:
Solve 2^(4x-3)=1/8, find the exact answer

Compute logarithm base 2 of both terms...

Kind regards

$\chi$ $\sigma$
 
Re: Solve

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

shorty888 said:
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

$\chi$ $\sigma$
 
Re: Solve

shorty888 said:
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
 

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