Laws of Logarithms

**Macleef** · December 9th 2007, 07:54 PM

Evaluate:

$log_28^{27} - log_464^{\frac {1}{5}}$

My work:

$= log_2(2^{3})^{27} - log_4 \sqrt [5] {64}$

$= log_2 2^{81} - \frac {1}{5} log_464$

$= 81 log_22 - \frac {1}{5} log_44^{3}$

And now what do I do? I don't know how to put both in the same base

**suissa** · December 9th 2007, 07:58 PM

next step:

81*(log2/log2) - (3/5)*(log4/log4)

You can do the rest.

I think thats it, if your initial reasoning is correct.

**DivideBy0** · December 9th 2007, 07:59 PM

Originally Posted by Macleef

Evaluate:

$log_28^{27} - log_464^{\frac {1}{5}}$

My work:

$= log_2(2^{3})^{27} - log_4 \sqrt [5] {64}$

$= log_2 2^{81} - \frac {1}{5} log_464$

$= 81 log_22 - \frac {1}{5} log_44^{3}$

And now what do I do? I don't know how to put both in the same base

You've done very well so far. All that's needed now is to realise that $\log_aa^b=b$ (convert back into exponential form to confirm this)

So $81\log_22^1-\frac{1}{5}\log_44^3=81\times 1 - \frac{1}{5} \times 3=\frac{402}{5}$

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