Solving this equation with exponents

[gregg0]

I have the following

\(\displaystyle 5n^2 (\frac{(\frac{1}{2}^logn - 1}{\frac{1}{2} - 1}) + logn\)

This is not rendered properly, it should be 1/2^logn - 1 on the numerator. I'm confused about how to solve the part in parenthesis.

 

[MarkFL]

You have given an expression rather than an equation. I gather you are attempting to simplfy:

\(\displaystyle 5n^2\left(\dfrac{\left(\frac{1}{2} \right)^{\log(n)}-1}{\frac{1}{2}-1} \right)+\log(n)\) ?

 

[gregg0]

Yes, I am trying to simply it.

 

[gregg0]

No, the problem is 1/2^logn - 1 which I don't know how to solve.

 

[mmm4444bot]

Are you familiar with these properties of exponents?

1/2 = 2^(-1)

(a^b)^c = a^(bc)

We can apply them to (1/2)^log(n) to get a power of 2.

The denominator 1/2 - 1 you know how to simplify, yes?

Then change from division by a fraction to multiplication by its reciprocal.

We'll have two powers of 2 that could be multiplied.

Do you know the property for multiplying two powers of the same base?