could someone help me with a question please?

Question

superhelp101 · Answer

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superhelp101 · Answer

@TuringTest Could you help me please? thanks

turingtest · Answer

is that$$\large4^{2x}=6^{x+1}$$?

superhelp101 · Answer

yes!

turingtest · Answer

well i guess we should just take the log of both sides first
base doesn't really matter

superhelp101 · Answer

yeah that is what i don't get. 
like how would i right that?

superhelp101 · Answer

write*

turingtest · Answer

well if you take the log of both sides you have something like$$\log(4^{2x})=\log(6^{x+1})$$you can then use the exponent rule for laws to simplify it
I should mention that I haven't solved this yet completely, so I should make sure I can.

turingtest · Answer

ah ok yeah i got it

superhelp101 · Answer

okay.
are both sides suppose to equal each other?

turingtest · Answer

well sure, everything we have done to one side we also did to the other, and they were equal before, so they still are

superhelp101 · Answer

alright. what should i do first?

turingtest · Answer

take the log of both sides, then use$$\log(x^a)=a\log x$$to simplify

superhelp101 · Answer

i still not completely getting it. sorry
could you show?

superhelp101 · Answer

me^

turingtest · Answer

do you understand up to here:$$\log(4^{2x})=\log(6^{x+1})$$?

superhelp101 · Answer

yes

superhelp101 · Answer

oh and btw my teacher got the answer 3.8188
but i don't see why

turingtest · Answer

if the problem is $$\large4^{2x}=6^{x+1}$$then the answer is not 3.8188

superhelp101 · Answer

ik but that is what my teacher got...

turingtest · Answer

i know for a fact that is not the answer, because wolfram and I both got the same answer, and it wasn't 3.8

superhelp101 · Answer

and i also plugged it in the calculator and it is wrong.
i don't see how he got that