Solve.

(1/4)^x+4 = √8

x+4 log 1/4= Log √8

I dont really no what to do from there...

Printable View

- Dec 16th 2009, 01:06 PMSkozLogarithms
Solve.

(1/4)^x+4 = √8

x+4 log 1/4= Log √8

I dont really no what to do from there... - Dec 16th 2009, 01:17 PMe^(i*pi)
$\displaystyle \left(\frac{1}{4}\right)^{x+4} = \sqrt{8}$

Note that

$\displaystyle \frac{1}{4} = 2^{-2}$

$\displaystyle \sqrt{8} = 2\sqrt{2} = 2^{1.5}$

Now you can convert all your numbers into base 2

$\displaystyle 2^{-2(x+4)} = 2^{1.5}$

Now if the bases are the same the exponents must also be the same

$\displaystyle -2x-8=1.5$

Solve the linear equation for x

I get x = -4.75 - Dec 16th 2009, 01:39 PMbigwave
well what was previously shown is much better way to solve it

however if you need to solve with logarothms

$\displaystyle \left(x+4\right)\left(log1 - log4\right) = \frac{log8}{2}$

$\displaystyle

-xlog4 - 4log4 = \frac{log8}{2}

$

$\displaystyle

-xlog4 = \frac{log8 + 8log4}{2log4} = -4.75

$

have assumed some of the steps