Melody

Excellent answer anonymous.  

Why don't you join up ?   It is super easy.  

So basically a log is a power (index or exponent)

Almost whenever you need to solve an equation where the unknown is an exponent then you will need to use logarithms.     

You may be interested to work through this unit.   

Ok Rosala I have looked at this question for you.

Anonymous's answer is incorrect.  It was correct up to the last line but then I think the answer was read off the calculator incorrectly.

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You answer is correct except that you have introduced a little error by saying that pi=3.14

This is just a 2 decimal place approximation for pi.  if you had used pi from the calculator then your answer would not have been 20cm. It would have been 19.0999cm correct to 4 decimal places.  OR 19cm correct to the nearest whole number.

Anyway the only error in your solution is rounding error.

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So yes I will certainly give your well earned 3 points.  CPhill must have given you the original 3 points or I would not be able to do this.

This is too large a number for most calcs to handle.

But a good approximation can be found this way.

$$8255^{33}\\\\
=(8255^{11})^3 \\\\
\approx (1.213075562 \times 10^{43})^3\\\\
\approx(1.213075562 \times 10^{43})^3\\\\
\approx(1.213075562^3) \times 10^{43*3}\\\\
\approx 1.785104155\times 10^{129}\\\\$$

$$3\sqrt{180}+\sqrt[2]{5}\\\\
=3\sqrt{36*5}+\sqrt{5}\\\\
=3*6\sqrt{5}+\sqrt{5}\\\\
=18\sqrt{5}+\sqrt{5}\\\\
=19\sqrt{5}\\\\$$

Thank you novice, that is a great answer.  

$$\\\dfrac{(r^{1/4}s^{2/5})^{20}}{r^2}\\\\
=\dfrac{(r^5s^8)}{r^2}\\\\
=r^3s^8$$

I didn't use the word "close". I said I communicate with admin.  Lol

She is Queen Guinevere from the British TV series "Merlin".

Don't you yanks know anything.  Lol

Fri 18/7/14

♬                                        ♬ ♬ MELODY ♬ ♬

Middle high students could find this interesting (Thanks NinjaDevo)