# Logarithms/intro to logarithms (how did they the get solution?)

1. Aug 1, 2012

### supernova1203

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

I have no idea how they came up with 7/2 as a solution...can anyone tell me how? Im not sure how logarithms and square roots work....how did they get the 1/2 exponent over the second 2? What is the train of thought required to solve a question like this??.... for example

you do this...then you do that...and using that you get 7/2...can someone explain to me? in detail how they got 7/2?

Thanks :)

Last edited: Aug 1, 2012
2. Aug 1, 2012

### HallsofIvy

Re: Logarithms/intro to logarithms

3. Aug 1, 2012

### supernova1203

Re: Logarithms/intro to logarithms

Try again...i re did them...and they seem to work now.

4. Aug 1, 2012

### HallsofIvy

Looks like magic!

Do you know the "properties of exponents"? $a^xa^y= a^{x+y}$ and $(a^x)^y= a^{xy}$

The logarithm is the opposite of the exponential: $log_a(a^x)= x$ and $a^{log_a(x)}= x$ (the technical term is "inverse") so logarithms have the opposite properties: $log_a(xy)= log_a(x)+ log_a(y)$ and $log_a(x^y)= y log_a(x)$

So, for example, $log_3(27\sqrt{3})= log_3(3^3(3^{1/2})= log_3(3^{3+ 1/2})= log_3(3^{7/2})= 7/2$
That, as well as, $log_2(8\sqrt{2})= log_2(2^3(2^{1/2})= log_2(2^{3+ 1/2})= log_2(2^{7/2})= 7/2$
are both applications of $log_a(a^x)= x$. The "logarithm" and "exponential" are inverse functions so they "cancel".

5. Aug 1, 2012

### Villyer

A notation that is used for square roots is to raise the number to the one-half power.

So the square root of two is equivalent to $2^{\frac{1}{2}}$.
On a similar note, the cube root of two is equivalent to $2^{\frac{1}{3}}$. (this isn't relevant for this problem, but I feel like it helps you understand the rule a little better.