Logarithms and exponential expressions

SourPatchParent

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Apr 19, 2015
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Firstly I am being asked to use the law of logarithms to simplify

y=1/4log(base 2)(8x-56)^16-12

am I on the right track thinking that I should start by multiplying the exponent 1/4 that is placed at the front with 16 to get the exponent 4? I'm quite lost here.


Secondly I have to solve for x- in this equation..

3^(x+4) - 5(3^x) = 684

If anyone can get me on the right track, that would be great.
 
Firstly I am being asked to use the law of logarithms to simplify
y=1/4log(base 2)(8x-56)^16-12
am I on the right track thinking that I should start by multiplying the exponent 1/4 that is placed at the front with 16 to get the exponent 4?
I have no idea what that is all about. I cannot read that messy notation.

Secondly I have to solve for x- in this equation..
3^(x+4) - 5(3^x) = 684
\(\displaystyle 3^{x+4}=3^4(3^x)=81(3^x)\) so we have
\(\displaystyle 76(3^x)=684\\3^x=9\\x=~?\)
 
Please hep me find out how to identify the period in graph :D

I need help D;
 
Firstly I am being asked to use the law of logarithms to simplify

y=1/4log(base 2)(8x-56)^16-12

am I on the right track thinking that I should start by multiplying the exponent 1/4 that is placed at the front with 16 to get the exponent 4? I'm quite lost here.
so y=log(8x-56)^4 -12 = log(8x-56)^4-log2^12 =log(8x-56)^4 - log(2^3)^4 = log [(8x-56)/8]^4 = log(x-7)^4 or 4log(x-7). NOTE: all logs are base 2
 
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