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It’s been suggested to me that I can solve for x in $5^x$$^-$$^1$$ = 2$ by converting to log base 10. I’ve tried converting to base 10 but I arrive at $x=\frac{\log2}{\log 5}+1$ and am unable to go further without using a calculator.

I’ve also tried to solve for x by first converting to base 5. The outcome $x=log_5^2+1$ is a little better, but I’d still need to resort to a calculator to solve for x.

Is there a way to do this that I have not considered? I had a look at this post (entitled Dividing logs with same base) but it didn't answer my question.

duckegg
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  • Hint: $2 = 10/5$. But yeah, you'll still need a calculator. This number is transcendental. – Brian Tung Jun 25 '22 at 07:59
  • "Converting to base $10$" is an awkward thing to write, since the numbers are already in base $10$ and there is no other base given. – Thomas Andrews Jun 25 '22 at 08:06
  • Also, we usually write $\log 2$ rather than $\log^2.$ Unless you mean something different? – Thomas Andrews Jun 25 '22 at 08:07
  • @BrianTung i’m not sure what to do with the hint. In my original transform, I’ve already converted the Right Hand Side to log base 10. – duckegg Jun 25 '22 at 08:12
  • Thanks for pointing that out to me @ThomasAndrews, I meant to write “log base 10”. I’ve corrected my post. – duckegg Jun 25 '22 at 08:13
  • @duckegg: I just meant that one can multiply both sides by $5$ to get $5^x = 10$, and then $x = \log_5 10 = \frac{1}{\log_10 5}$. But as I said, that number is transcendental and can't be simplified further. Were you expecting a different form? – Brian Tung Jun 25 '22 at 16:34

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