question about laws of logarithms

**Jones** · January 1st 2010, 02:27 AM

Hi,

The laws of logarithms are pretty confusing to me.

I have the following problem:

$\frac{1}{2^x} = \frac{5}{8^{x+3}}$

This can be rewritten as $2^{-x} = 5*8^{-(x+3)}$

if you log both sides you'll get: $-xln2 = ln5*-(x+3)ln8$

so when we have a multiplication like this -(x+3)*ln8 does this imply addition?

I don't understand the basic arithmetic operations seem to have been switched, just to maximize confusion.

Someone please explain how this works =/

Jones

**mr fantastic** · January 1st 2010, 02:51 AM

Originally Posted by Jones

Hi,

The laws of logarithms are pretty confusing to me.

I have the following problem:

$\frac{1}{2^x} = \frac{5}{8^{x+3}}$

This can be rewritten as $2^{-x} = 5*8^{-(x+3)}$

if you log both sides you'll get: $-xln2 = ln5*-(x+3)ln8$

so when we have a multiplication like this -(x+3)*ln8 does this imply addition?

I don't understand the basic arithmetic operations seem to have been switched, just to maximize confusion.

Someone please explain how this works =/

Jones

$\frac{1}{2^x} = \frac{5}{8^{x+3}}$

$\Rightarrow 2^{-x} \cdot 8^{x + 3} = 5$

$\Rightarrow 2^{-x} \cdot (2^3)^{x + 3} = 5$

$\Rightarrow 2^{-x} \cdot 2^{3x + 9} = 5$

$\Rightarrow 2^{2x + 9} = 5$

$\Rightarrow \log 2^{2x + 9} = \log 5$

$\Rightarrow (2x + 9) \log 2 = \log 5$

$\Rightarrow 2x + 9 = \frac{\log 5}{\log 2}$

and it is simple to make x the subject. The log is to any base you want it to be, by the way.

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