Question #070b2

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Answer 1 · 2016-11-23T00:40:47

Assuming that the logarithm is base 10:
${log}_{10} (9) = \frac{ln (9)}{ln (10)}$ $log_10(9) = ln(9)/ln(10)$

Answer 2 · 2017-08-18T20:00:43

${log}_{10} (9) = \frac{ln (9)}{ln (10)}$ $log_10(9) = (ln(9))/(ln(10))$

Consider ${log}_{10} (9)$ $log_10(9)$

Set ${log}_{10} (9) = x$ $" " log_(10)(9) = x$

Another way of writing this is: $10^{x} = 9$ $" "10^x=9$

Set $e^{b} = 9 \Rightarrow b = ln (9)$ $" "e^b=9=>color(red)(b=ln(9))$

But we have: $e^{b} = 9 = 10^{x}$ $" " e^b=9=10^x$

Take natural logs of both sides and using the principle that
$ln (10^{x})$ $ln(10^x)$ is the same as $x ln (10)$ $xln(10)$

$b ln (e) = x ln (10)$ $bln(e)=xln(10)$

but $ln (e) = 1$ $ln(e)=1$

$b = x ln (10)$ $b=xln(10)$

Thus $x = \frac{b}{ln (10)}$ $x=(color(red)(b))/(ln(10))$

but $b = ln (9)$ $b=ln(9)$ giving

$x = \frac{ln (9)}{ln (10)}$ $x=ln(9)/(ln(10))$

But we set ${log}_{10} (9) = x$ $log_(10)(9) = x$ giving:

${log}_{10} (9) = \frac{ln (9)}{ln (10)}$ $log_10(9) = (ln(9))/(ln(10))$