rewriting logs help

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jimmypop

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Apr 26, 2009
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I have a series of study questions like this and I am at a bit of a loss?

Rewrite the expression as the sum and difference or product of logarithm

log (base 3) (3rd root of 3x-2)/5x^2
Here is an image of the problem in case I am not clear on the describing the expression..
123.tif

Any help would be appreciated.
 
\(\displaystyle \log_4\frac{\sqrt{a}}{b^5}=\log_4 \sqrt{a}-\log_4b^5\).
\(\displaystyle \log_4\sqrt{a} = \log_4a^\frac{1}{2}=\frac{1}{2}\log_4a\)
\(\displaystyle \log_3b^5=5\\log_3b\)
 
I apperently am messing up somewhere along the way?

I was told I went lost starting at this step......

log(base 3)(3x-2)^1/3 - 2log (base 3) 5^x

I guess this step is wrong? what am I missing?
 
jimmypop said:
I was told I went lost starting at this step......

log(base 3)(3x-2)^1/3 - 2log (base 3) 5^x
Click to expand...
The only way the "2" can go out in front is if it is on everything inside the log. Is it, or is it only on the "x"?

Also, there is no reason to turn to 5x^2 into a 5^(x^2). :oops:
 
jimmypop said:
I have a series of study questions like this and I am at a bit of a loss?

Rewrite the expression as the sum and difference or product of logarithm

log (base 3) (3rd root of 3x-2)/5x^2
Here is an image of the problem in case I am not clear on the describing the expression..
123.tif

Any help would be appreciated.
Click to expand...

\(\displaystyle Log_3\frac{(3x-2)^{\frac{1}{3}}}{5x^2}\)

\(\displaystyle = Log_3(3x-2)^{\frac{1}{3}} - Log_3(5x^2)\)

\(\displaystyle = Log_3(3x-2)^{\frac{1}{3}} - [Log_3(5) + Log_3(x^2)]\)

Now continue.....
 
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