What is the Solution for $5^x+2=12$?

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Discussion Overview

The discussion revolves around solving the equation $5^x + 2 = 12$ and a potential confusion regarding whether the equation should be $5^x + 2 = 126$. Participants explore the solution process and clarify the problem statement.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a solution for $5^x + 2 = 12$, arriving at $x = \frac{\ln(10)}{\ln(5)} \approx 1.4307$.
  • Another participant suggests that using base 5 logarithm ($\log_5$) could be an improvement but acknowledges that most calculators do not support this directly.
  • There is a question raised about the accuracy of the problem statement, with one participant noting a discrepancy between the thread title and the equation solved.
  • If the equation is actually $5^x + 2 = 126$, a participant provides a solution indicating that $x$ would be slightly less than 3, specifically $x \approx 2.99500933$.
  • One participant apologizes for editing the title to reflect the equation they solved, clarifying the confusion.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the correct equation to solve, with some focusing on $5^x + 2 = 12$ and others considering $5^x + 2 = 126$. The discussion remains unresolved as to which equation is the intended problem.

Contextual Notes

The discussion highlights potential typos and the importance of clarity in problem statements. There is also a mention of calculator limitations regarding logarithmic functions.

karush
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Solve $5^x+2=12$
$5^x=10\implies x\ln(5)=\ln(10)\implies x=\dfrac{\ln(10)}{\ln(5)}\approx \boxed{1.4307}$

ok i think this is ok but typos or better steps maybe
 
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karush said:
Solve $5^x+2=12$
$5^x=10\implies x\ln(5)=\ln(10)\implies x=\dfrac{\ln(10)}{\ln(5)}\approx \boxed{1.4307}$

ok i think this is ok but typos or better steps maybe
Looks good. The only possible improvement I might suggest is to use [math]log_5[/math] instead of ln (then [math]x = log_5(10)[/math]) but in practical terms most calculators don't have a key to evaluate [math]log_5(10)[/math] so I'd use ln, too.

-Dan
 
You solved $5^x+ 2= 12$ but the title of this thread was "Solve $5^x+ 2= 126$". Which is it?
 
If the problem is actually $5^x+ 2= 126$ then $5^x= 124$, $log(5^x)= x log(5)= log(124)$ so $x= \frac{log(124)}{log(5)}$. Since $5^3= 125$ x will be a little less than 3. Using a calculator, x is about 2.99500933 .
 
County Boy said:
You solved $5^x+ 2= 12$ but the title of this thread was "Solve $5^x+ 2= 126$". Which is it?
sorry I edited the title to what I solved
 

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