Any help?
solve 10=3^x+5 -2

Question

Any help?
solve 10=3^x+5 -2

whpalmer4 · Answer

That equation does not look reasonable.  Are you sure you have copied it correctly?

jellybot23 · Answer

yes, it is 10=3^x+5 -2
The -2 is separate from the exponent

whpalmer4 · Answer

you can't write it like that if you mean something other than
$$10 = 3^x + 5 -2$$

whpalmer4 · Answer

Is it supposed to be $$10 = 3^{x+5}-2$$

jellybot23 · Answer

yes! the 2nd one :)

whpalmer4 · Answer

If that's what you intended, you need to put parentheses around the exponent, like this:
10 = 3^(x+5) - 2

remember the order of precedence:  PEMDAS

parentheses
exponentiation
multiplication
division
addition
subtraction

so 
10 = 3^x + 5 - 2

first operation done is 3^x, then you add 5, then you subtract 2

jellybot23 · Answer

oh okay. Then I meant 10=3^(x+5) -2
Sorry about that

whpalmer4 · Answer

usually, if you copy some equation where there are exponents or anything that relies on positioning, like a complicated fraction such as
$$\frac{a+b}{c+d}$$you need to use some parentheses to make it clear (or the equation editor button).

a lot of people might right that fraction as 
a+b/c + d

but that really means $$a + \frac{b}{c} +d $$as they have written...

whpalmer4 · Answer

but now that we have that out of the way, on to your problem :-)

$$10 = 3^{x+5} - 2$$what do you think a good first step might be?

jellybot23 · Answer

Okay, I will remember that :)
And, well I thought you had to do $$\log_{3}10 = x+5$$ first

whpalmer4 · Answer

wait, what happened to the -2?

jellybot23 · Answer

I was going to subtract two after I got my answer. I'm not very good at logs :(  They always confuse me

whpalmer4 · Answer

Can't do that.  But even if you could, why wouldn't you get rid of that 2 first by adding 2 to both sides and simplifying the field of battle?

jellybot23 · Answer

Oh true! so 12=3^(x+5)

jellybot23 · Answer

Then do I do $$\log_{3}12=x+5 $$

whpalmer4 · Answer

yes.  keep going...

jellybot23 · Answer

I plugged it into my calculator and got approximately -2.738

whpalmer4 · Answer

And if you plug that value into the right hand side of the original equation, do you get a number very close to 10?

jellybot23 · Answer

yes! :)

whpalmer4 · Answer

Excellent!

So, I agree that logs can be confusing, but if you think of them as the reverse of exponentiation, that helps me, at least.

jellybot23 · Answer

Awesome! Thank you so so much :D

whpalmer4 · Answer

you're welcome!