N00b here and i got a question (logarithms)

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    Logarithms
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Discussion Overview

The discussion revolves around solving an exponential equation involving logarithms, specifically the equation 3^(2x) = 80. Participants seek clarification on the algebraic steps required to solve the equation and approximate the result.

Discussion Character

  • Homework-related

Main Points Raised

  • One participant presents the equation 3^(2x) = 80 and requests assistance in solving it.
  • Another participant outlines a step-by-step approach to solving the equation using natural logarithms, applying the property of logarithms to isolate x.
  • A third participant expresses gratitude and inquires about formatting numbers in a specific way in the forum.
  • A fourth participant provides guidance on how to format text in the forum and references another thread for further assistance.

Areas of Agreement / Disagreement

There is no disagreement present; participants are collaboratively discussing the solution process and formatting questions.

Contextual Notes

The discussion does not address potential limitations or assumptions in the mathematical steps provided.

Who May Find This Useful

Individuals seeking help with logarithmic equations or those interested in formatting posts on the forum may find this discussion beneficial.

mathdummy
Okay.
"Solve the exponential equation with base a algebraically. Approximate the result to three decimal places."

3^2x=80

(fyi, the answer is ln80/2ln3 = 1.994)

Don't understand how to solve, please help!
 
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As you already have the answer and you want to understand the question, I think it's okay to take you through it step-by-step:

32x = 80

take the natural logarithm of both sides:

ln 32x = ln 80

use the rule ln xy = yln x:

2xln 3 = ln 80

2x = ln 80/ln 3

x = ln 80/2ln 3
 
Ah. Thanks a lot! For future reference, how can i show the numbers being squared in the corner and small like you were able to
 

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