Solve for x: 4^(x-5) = 7^(2x-1) | Equation Help

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Discussion Overview

The discussion revolves around solving the equation ${4}^{x-5} = {7}^{2x-1}$. Participants explore methods for manipulating the equation, particularly through the use of logarithms.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant requests help with the equation ${4}^{x-5} = {7}^{2x-1}$.
  • Another participant suggests taking the logarithm of both sides, leading to the equation $(x-5)\log\, 4 = (2x-1) \log\, 7$, and notes that it can be treated as a linear equation.
  • Several participants ask for clarification on how the logarithmic transformation was derived, specifically referencing the property $\log\,a^x = x \log\,a$.
  • One participant indicates they have understood the explanation provided.

Areas of Agreement / Disagreement

There is no explicit consensus on the solution process, as participants are still clarifying steps and understanding the logarithmic transformation.

Contextual Notes

The discussion does not resolve the equation or provide a final solution, and participants have not confirmed the correctness of the steps taken.

Nikolas7
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Help me with the following equation:
${4}^{x-5}$=${7}^{2x-1}$
 
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Nikolas7 said:
Help me with the following equation:
${4}^{x-5}$=${7}^{2x-1}$

you can take log on both sides and get $(x-5)\log\, 4 = (2x-1) log\, 7$ and because it is linear equation you can proceed to finish it
 
Please, show details how you got (x−5)log4=(2x−1)log7
 
Nikolas7 said:
Please, show details how you got (x−5)log4=(2x−1)log7

this is as per definition $\log\,a ^ x = x \log\, a $
 
Yes, I figured out, thanks
 

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