DewdropDancer

OK! Thank you!

The prime factorization of 90 is 3 * 3 * 5 * 2 = 90

Same thing as hihihi and KSW said... but more detailed (i think)

\(2n+5=15\:\:\)

\(\mathrm{Subtract\:}5\mathrm{\:from\:both\:sides}\)

\(2n+5-5=15-5\)

\(\mathrm{Simplify}\)

\(2n=10\)

\(\mathrm{Divide\:both\:sides\:by\:}2\)

\(\frac{2n}{2}=\frac{10}{2}\)

\(\mathrm{Simplify}\)

\(n=5\)

Vivek, I believe we have the same answer, but anyways, Guest could you specify where you think I got the problem wrong?

I believe your question is incomplete, before you post a question, reread what you wrote so that we understand what you are asking.

\(2^x+2^{x+1}+2^{x+2}+2^{x+3}=\frac{15}{2}\)

\(15\cdot \:2^x=\frac{15}{2} \)
\(\frac{15\cdot \:2^x}{15}=\frac{\frac{15}{2}}{15}\)

\(2^x=\frac{1}{2}\)

\(x=-1\)

I am not sure why this picture is blocked permanently, is there a way you can fix it so I can help you with this problem?

The Hexagon has 18 triangles, which each have 1/3 area, and there are 12 triangles in the light blue star, so 

\(12*\frac{1}{3}=4\)

\(\begin{bmatrix}y=\log _2\left(x\right)\\ 2x+y-5=0\end{bmatrix}\)

\(\begin{bmatrix}2x+\log _2\left(x\right)-5=0\end{bmatrix}\)

\(y=\log _2\left(2\right)\)

\(y=1,\:x=2\)