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

[supernova1203]

Homework Statement

Homework Equations

The Attempt at a Solution

I have no idea how they came up with 7/2 as a solution...can anyone tell me how? I am 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 :)

 

[HallsofIvy]

Your links don't work.

 

[supernova1203]

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

 

[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".

 

[Villyer]

how did they get the 1/2 exponent over the second 2?

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.