e^2x= e((x+log5)/loge)-6

Question

terenzreignz · Answer

lol... don't you just love how they complicate the simple? ^_^

terenzreignz · Answer

$$\Large e^{2x}= e^{\frac{x+\log 5}{\log e}}-6 $$
Is this it?

anonymous · Answer

yes!

terenzreignz · Answer

Well this is a bit hard, isn't it?
Perhaps we can use the change of base formula somewhere... do you know the change of base formula?

terenzreignz · Answer

Also, are you sure it's not like THIS:
$$\Large e^{2x}= e^{x + \frac{\log 5}{\log e}}-6$$

anonymous · Answer

$$\frac{ 1 }{ \log_{b}  a }$$ right?

anonymous · Answer

Maybe.. the TA posted the problems and then sent an email saying there was a typo.

terenzreignz · Answer

Well, you better confirm, because this can be the difference between a simple or outrageously and impossibly complicated answer.

anonymous · Answer

he takes forever to respond. I'll copy and paste his correction he sent.

terenzreignz · Answer

You do that :)

anonymous · Answer

the exponent should be x + log(5)/log(e), rather than x log(5)/log(e).<< this is what he emailed us.

terenzreignz · Answer

Haha... my instincts serve me correctly, don't they? :P

anonymous · Answer

here is the original email

terenzreignz · Answer

yeah, so it really is
$$\huge e^{2x}= e^{x+\frac{\log 5}{\log e}}-6$$

anonymous · Answer

okay whew... I tried it on mathway the other way and it said no answer.. I was getting aggrivated.

anonymous · Answer

so how do I even begin this problem.. I've looked through our book and haven't seen anything like it

terenzreignz · Answer

hang on... lag

terenzreignz · Answer

First, what is $$\Large \frac{\log 5}{\log e}$$