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

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Discussion Overview

The discussion revolves around solving the exponential equation \(2^{(4x-3)} = \frac{1}{8}\). Participants explore different methods to find the exact solution, including logarithmic approaches and direct comparison of exponents.

Discussion Character

  • Homework-related

Main Points Raised

  • One participant suggests rewriting \(1/8\) as \(2^{-3}\) to facilitate solving the equation by equating the exponents.
  • Another participant proposes using logarithms, specifically logarithm base 2, to solve the equation.
  • Several participants express confusion about the steps involved in solving the equation and seek clarification.
  • A participant reiterates the method of equating the exponents, stating that if \(a = b\), then \(\log_{2} a = \log_{2} b\) can be applied.
  • Another participant emphasizes the importance of quoting previous posts for clarity in responses.

Areas of Agreement / Disagreement

Participants generally agree on the methods to approach the problem, but there remains uncertainty among some about the execution of these methods. The discussion does not reach a consensus on the solution process, as confusion persists.

Contextual Notes

Some participants have not fully grasped the logarithmic approach or the implications of equating exponents, indicating a potential gap in understanding foundational concepts.

shorty888
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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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