Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

How do you solve $ln x = 7.25$ $lnx=7.25$ ?

2 Answers

Jul 15, 2016

$x = e^{7.25} \approx 1408.105$ $x=e^7.25~~1408.105$

Explanation:

Remember from basic definitions
$XXX ln (a) = c$ $color(white)("XXX")ln(a)=c$ means $e^{c} = a$ $e^c=a$

Therefore if
$XXX ln (x) = 7.25$ $color(white)("XXX")ln(x)=7.25$
then
$XXX x = e^{7.25}$ $color(white)("XXX")x=e^7.25$
(using a calculator we can find the approximation $e^{7.25} = 1408.105$ $e^7.25=1408.105$ )

Jul 15, 2016

$e^{7.25}$ $e^7.25$ , or $10^{7.25}$ $10^7.25$ , or $b^{7.25}$ $b^7.25$ , $b$ $b$ being the base of the logarithm.

Explanation:

You can only use the fact that the logarithm and the exponential are one the inverse function of the other. This means that

$e^{ln x} = x$ $e^lnx=x$ and $ln (e^{x}) = x$ $ln(e^x)=x$

Using this property, you can put both left and right member at the exponent:

$ln (x) = 7.25 \Leftrightarrow e^{ln (x)} = e^{7.25}$ $ln(x)=7.25 \iff e^ln(x)=e^7.25$

But we have just observed that $e^{ln (x)} = x$ $e^ln(x)=x$ , so we have

$x = e^{7.25}$ $x=e^7.25$ .

This, of course, assuming that by "log" you meant the natural one. If, for example, you use base 10 logarithm, you should change $e$ $e$ with $10$ $10$ , like this:

$log (x) = 7.25 \Leftrightarrow 10^{log (x)} = 10^{7.25} \Leftrightarrow x = 10^{7.25}$ $log(x)=7.25 \iff 10^log(x)=10^7.25 \iff x=10^7.25$

Related questions

See all questions in Logarithmic Models

Impact of this question

11554 views around the world

You can reuse this answer
Creative Commons License