Question

How do I add these logarithms? Help, pretty please?

Answer 1

\[\log _{3}(x+7) + \log _{3}3 = 2\]

Answer 2

I know I can make it a single log by making it\[\log _{3}3(x+7) = 2\]

Answer 3

That's one way of starting off the problem, but the quickest way I saw was at first that
\[\log_{3} 3 = 1\]

Answer 4

So from that you'd just have
\[\log_{3}(x+7) \]

Answer 5

What do you think would be the next step?

Answer 6

Srry i meant:\[\log_{3}(x+7) = 1 \]

Answer 7

Hmmm.. solving for x? or... moving x over to the other side? I'm not sure...

Answer 8

Alright yea, so to solve for x, we would have to exponentiate both sides.

Answer 9

What I mean by exponentiate is just this:\[\log_{a}b = x \]\[a^x = b\]

Answer 10

I'm a little bit confused on how to do that.. like which numbers to put where in order to exponentiate.

Answer 11

So in this situation, all that needs to be done is to just perform the base of the logarithm (3) to both sides of the equation.\[3^{\log_{3}(x+7)} = 3^{1}\]

Answer 12

Since exponentiation is the inverse of the logarithm function, it undos it. So\[3^{\log_{3}(x+7)} = x+7\]

Answer 13

So, do the (x + 7)s cancel each other out, or..?

Answer 14

I'm sorry I was just showing the one side of the equation. The equation as I wrote it was:\[3^{\log_{3}(x+7)} = 3\]\[x+7 = 3\]

Answer 15

Ohh! Okay, so if I solve for x it would be -4?

Answer 16

Yes! The whole problem there was to do the inverse operation on both sides of the equation so as to isolate x in a form that can be manipulated easily. The way that you proposed at first would have gone like so:
\[\log_{3}(x+7) + \log_{3}(3) = 2\]\[\log_{3}(3(x+7)) = 2\]\[3^{\log_{3}(3(x+7))} = 3^{2}\]\[3x + 21 = 9\]\[3x = -12\]\[x = -4\] But it's just a little more work. Either way gets the answer though so It usually doesn't matter.

Answer 17

Okay!! Thank you very much!

Answer 18

Np niggie.