# Math Help - [SOLVED] Weird Logarithm Question

1. ## [SOLVED] Weird Logarithm Question

y^5.6 = 4

Solve for y using logarithms.

I know how to solve this question without logs, and I know how to solve logs when the variable is in the exponent.

2. You have to remember the rules for logs for this question:

$y^{5.6} = 4$
$\ln y^{5.6} = \ln 4$
$5.6\ln y = \ln 4$
$\ln y = \frac{\ln 4}{5.6}$
$y = e^{\frac{\ln 4}{5.6}}$
$y = 1.28....$

3. Originally Posted by Some_One
y^5.6 = 4

Solve for y using logarithms.
I'm assuming this is $y^{5.6}=4$ and that the period isn't used to represent multiplication.

The solution depends on the use of a property of logarithms. I suggest you brush up on the various properties, but the one needed here is that $\log x^z=z\log x\text.$

So, we start by taking the log of both sides, and then we can apply the property:

$y^{5.6}=4\Rightarrow\ln y^{5.6}=\ln4$

$\Rightarrow5.6\ln y=\ln4\Rightarrow\ln y=\frac{\ln4}{5.6}$

Now, exponentiating both sides,

$e^{\ln y}=e^{(\ln4)/5.6}\Rightarrow y=\left(e^{\ln4}\right)^{1/5.6}$

$\Rightarrow y = 4^{1/5.6} = 4^{5/28} = 2^{5/14}\approx1.2809$

I took the natural log (base $e$) of both sides, but you could use any base.