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## LaronJaxon Group Title What is the solution to the equation 1/sqrt 8= 4^(m - 3) ? one year ago one year ago

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1. zordoloom Group Title

m=9/4

2. terenzreignz Group Title

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

3. terenzreignz Group Title

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

4. terenzreignz Group Title

$\huge \frac{1}{2^{\frac{3}{2}}}$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. terenzreignz Group Title

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

6. terenzreignz Group Title

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

7. terenzreignz Group Title

And the rest, is history :D