1

How do I solve? $$ x = \log_{10} 5$$

Until I understood until for now, it's same as:

$$ 10^x = 5 $$

and $x$ will be a value $> 0$ and $< 1$ because if it's $1$, the value $= 10$

But someone is solving this with $$ y = 10^5 $$

By using this last method, don't give the result as my book or calculator. I'm looking for learn by using the first, how to do this?

martini
  • 84,101
Assembly
  • 87
  • 4
    Don't listen to that someone. You are right. – njguliyev Oct 09 '13 at 07:07
  • You don't “solve”: $x=\log_{10}5$ is already finished. What else you can do is computing an approximation of the number: $\log_{10}5\approx0.69897$, but it's an irrational number. – egreg Oct 09 '13 at 07:10

2 Answers2

2

What do you mean by "solve"?

If you mean to find the value of $x$ that makes that true, then the problem is easy: that value is $\log_{10} 5$.

If you mean to find a decimal expression that is approximately equal to the value of $x$ that makes that true, then there isn't really anything to do more than use your calculator. There are ways to compute logarithms by hand, but there is little practical use for such, and doing so is surely beyond the scope of your course.

1

$ \log_{10} 5$ literally means "what power of 10 equals 5?" which you have correctly written as
$10^x = 5$. There is no way to find x without looking into your log-tables.

rnjai
  • 1,774