I dont know how to enter 2 ln5+ 2 ln3 into my calculator!! please help!

Question

whpalmer4 · Answer

What kind of calculator do you have?

anonymous · Answer

TI30xa

anonymous · Answer

or know how to write it another way so i can type it in differently?

whpalmer4 · Answer

Looks like you should be able to enter it just as it looks: 2 LN 5 + 2 LN 3
When it doubt, you could use parentheses.

whpalmer4 · Answer

What do you get when you try to enter it?

anonymous · Answer

I get 8 but i have to do more with the problem and everytime enter the answer in my math homework online it says its wrong

whpalmer4 · Answer

Hmm, you're taking the natural log of 5 and 3?  That's not going to give you such a nice round number.

anonymous · Answer

the problem is 2^2-x=3^5x+7

whpalmer4 · Answer

What if you just try to take the natural log of 5, what do you get?

whpalmer4 · Answer

$$2^2 - x = 3^{5x} + 7$$?

anonymous · Answer

its 2 to the power of 2-x = 3 to the power of 5x+7

whpalmer4 · Answer

$$2^{2-x} = 3^{5x+7}$$
?

anonymous · Answer

yes

whpalmer4 · Answer

It's a good news/bad news kind of situation.  Good news is that you don't have to evaluate that expression you asked about. Bad news is that means you haven't done the problem correctly :-)

anonymous · Answer

lol how do I do it?

whpalmer4 · Answer

What if you multiply both sides by $2^{2-x}$?

whpalmer4 · Answer

Then you can use a trick: $$e^{\log x} = x$$ and $$e^0 = 1$$

whpalmer4 · Answer

Sorry, I mistyped my first suggestion: $2^{x-2}$ is what you want to multiply both sides with.

anonymous · Answer

so $$3^{5x+7}\times2^{2-x}$$

whpalmer4 · Answer

That gives us
$$2^{2-x}*2^{x-2} = 2^{x-2}3^{5x+7}$$Adding the exponents on the left (common base)
$$1 = 2^{x-2}3^{5x+7}$$

whpalmer4 · Answer

But with the trick I mentioned, we can write that as 
$$1 = e^{\log 2^{x-2}} e^{\log 3^{5x+7}}$$

whpalmer4 · Answer

Now we can use the property that $$\log a^n = n \log a$$and write it as
$$1 = e^{(x-2)\ln 2}e^{(5x+7)\ln 3}$$and then add the exponents of a common base to get
$$1 = e^{(x-2)\ln 2 + (5x+7)\ln 3}$$Do you think you can take it from there?

anonymous · Answer

yes thank you!

whpalmer4 · Answer

I need to go run some errands, but post your answer and I'll have a look when I return.

anonymous · Answer

ok