# Thread: Solve for x without log's

1. ## Solve for x without log's

Solve for x $ln\, x=1/5 ln(A-1) - 4 ln\,B +4$, expressed without the log's.

My attempt:
$
ln\, x = \frac{{1}}{5}ln\,(A-1)-4ln\, B+4 \\$

$ln\, x = ln\,(A-1)^{1/5}-ln\, B^{4}+4 \\$
$ln\, x = ln\,\left(\frac{(A-1)^{1/5}}{B^{4}}\right)+4 \\$
$e^{ln\, x} = e^{ln\,\left(\frac{(A-1)^{1/5}}{B^{4}}\right)+4} \\$
$x = \left(\frac{(A-1)^{1/5}}{B^{4}}\right)\times e^{4}$

Can this be done better; or condensed more?

Thanks!

2. Originally Posted by Jodles
Solve for x $ln\, x=1/5 ln(A-1) - 4 ln\,B +4$, expressed without the log's.

My attempt:
$
ln\, x = \frac{{1}}{5}ln\,(A-1)-4ln\, B+4 \\$

$ln\, x = ln\,(A-1)^{1/5}-ln\, B^{4}+4 \\$
$ln\, x = ln\,\left(\frac{(A-1)^{1/5}}{B^{4}}\right)+4 \\$
$e^{ln\, x} = e^{ln\,\left(\frac{(A-1)^{1/5}}{B^{4}}\right)+4} \\$
$x = \left(\frac{(A-1)^{1/5}}{B^{4}}\right)\times e^{4}$

Can this be done better; or condensed more?

Thanks!
From your second line if you say that $4 = \ln(e^4)$ then your second line becomes

$\ln(x) = \ln \left(\dfrac{\sqrt[5]{A-1} \cdot e^4}{B^4}\right)$

The continue as normal to get the same answer.

Your way is fine though, if anything quicker