OpenStudy (laronjaxon):

What is the solution to the equation 1/sqrt 8= 4^(m - 3) ?

5 years ago
OpenStudy (anonymous):

m=9/4

5 years ago
terenzreignz (terenzreignz):

\[\large \frac{1}{\sqrt{8}}=4^{m-3}\]

5 years ago
terenzreignz (terenzreignz):

One nifty way of doing this is to try to express both sides of the equation as exponentials, ie, one base, one exponent... preferably with the same base. So, let's start with \[\frac{1}{\sqrt{8}}\] 8 is just 2³, so let's put it that way...\[\large \frac{1}{\sqrt{2^{3}}}\] And remember that taking the square root means raising something to the 1/2 power, so...\[\huge \frac{1}{\left( 2^3 \right)^{\frac{1}{2}}}\] Using laws of exponents, you get

5 years ago
terenzreignz (terenzreignz):

\[\huge \frac{1}{2^{\frac{3}{2}}}\]Now remember that \[\large a^{-n}=\frac{1}{a^n}\] So eventually, we're left with \[\huge \frac{1}{\sqrt{8}} = 2^{-\frac{3}{2}}\] Now on to the other side of the equation...

5 years ago
terenzreignz (terenzreignz):

\[\large 4^{m-3}\] But 4 = 2² So, we can write it as \[\huge (2^2)^{m-3}\] Again, using laws of exponents, it is just equal to \[\huge 2^{2(m-3)}=2^{2m - 6}\]

5 years ago
terenzreignz (terenzreignz):

So, your problem becomes... \[\huge 2^{-\frac{3}{2}}=2^{2m - 6}\] Which can only mean \[\large -\frac{3}{2}=2m - 6\]

5 years ago
terenzreignz (terenzreignz):

And the rest, is history :D

5 years ago
OpenStudy (anonymous):

The answer is m=9/4 (just took the test and got it right)

3 years ago