Question

Solve the Logarithmic equation. Please show me step by step, thank you! (Equation below)

Answer 1

\[\log_{10}(x-30) = 3 - \log_{10}x \]

Answer 2

@Mertsj perhaps you could be of assistance?

Answer 3

Since the logs are of base 10, I'll just write log(x)
log(x - 30) = 3 - log(x)

Answer 4

Add log(x) to both sides

Answer 5

log(x - 30) + log(x) = 3

Answer 6

Do you know the law of the sum of logs?

Answer 7

That's where if they're added you change it to multiplication, and if it's subtraction you change it to division, correct?

Answer 8

Correct. What can you do to the left side now?

Answer 9

log(x - 30)/x ?

Answer 10

It's a sum of logs on the left, not a subtraction.

Answer 11

Oh shoot, right, my bad, so then:
log(x)(x - 30)

Answer 12

log(x^2 - 30x) = 3

Answer 13

Then I convert it to exponential?
(x^2 - 30x) = 10^3

Answer 14

Correct.

Answer 15

Now solve the quadratic equation for x.

Answer 16

Thank you very much!

Answer 17

wlcm

Answer 18

remember to check both solutions of the quadratic eq in the original logarithmic eq since the domain of the log function is restricted.

Answer 19

Cannot be negative log or log(0) correct?

Answer 20

Correct.
log(x) is defined for positive x only.
That does not mean that a negative value of x from the solution of the quadratic eq will produce an invalid argument for the log function. The best way to find out is to check both solutions.
In other words, a negative solution to the quadratic eq may still work. You need to check both solutions to be sure.

Answer 21

Okay. and since I've got your attention still, what would I do for the next step in this problem? It's the same kind, except it was subtraction, so it was changed to division. Still not good at these types of problems. :/

\[\log_{2} (x+2)/(x-5) = 1\]

Answer 22

You mean the log base 2 of the entire fraction? Or the log base 2 of (x + 2), then divided by x - 5?

Answer 23

Yes, log base 2 then in parenthesis the x+2 over x-5

Answer 24

OK, then change the log eq into an exponential eq

Answer 25

Ah, alright, got it. Thanks again!