Question

Why am I getting this question wrong?

Answer 1

You need to write out your first step for me. I need to see what you are doing.

Answer 2

\[\large (2x+2)Log4 = (x-3)Log5\]

Answer 3

correct, now what is the next thing you need to do?

Answer 4

take logarithm of both sides first, so \(\Large{ \bf 4^{2x+2}=5^{x-3}\qquad \textit{taking say log to both sides}\\ \quad \\
log(4^{2x+2})=log(5^{x-3})}\)
notice, now you can take out the exponents from the log, as shown above by cronus8992

Answer 5

yes he knows this is his problem...

Answer 6

ohh... smokes.. is @cronus8992 the fellow asking hheh, so much for my reading up =)

Answer 7

ok

Answer 8

;)

Answer 9

\(\bf 4^{2x+2}=5^{x-3}\qquad \textit{taking say log to both sides}\\ \quad \\
log(4^{2x+2})=log(5^{x-3})\implies (2x+2)log(4)=log(5)\\ \quad \\
2x+2=\cfrac{log(5)}{log(4)}\) and so on

Answer 10

hmm... so much for my typos =)

Answer 11

\(\bf 4^{2x+2}=5^{x-3}\qquad \textit{taking say log to both sides}\\ \quad \\
log(4^{2x+2})=log(5^{x-3})\implies (2x+2)log(4)=(x-3)log(5)\\ \quad \\
\cfrac{2x+2}{x-3}=\cfrac{log(5)}{log(4)}\)

Answer 12

hmm

Answer 13

\(\bf 4^{2x+2}=5^{x-3}\qquad \textit{taking say log to both sides}\\ \quad \\
log(4^{2x+2})=log(5^{x-3})\implies (2x+2)log(4)=(x-3)log(5)\\ \quad \\
2x(log(4))+2log(4)=xlog(5)-3log(5)\\ \quad \\
\implies 2x(log(4))-xlog(5)=-3log(5)-2log(4)\)

from there,take common factor, to get "x"