Question

Solve for x.
logx+log(x-15)=2

Answer 1

hello :)
what did you try??

tried combining those 2 logs?

Answer 2

I know what you have to do, however, I am not sure how to do it :(

Answer 3

I don't know how to combine the logs

Answer 4

\(\log A + \log B = \log AB \)

so how about log x + log (x-15)
?

Answer 5

is it 20?

Answer 6

you're asking if x = 20 ?

how did you do all the steps so fast! :O

Answer 7

I've been working on this problem for a really long time and I wasn't sure if I was doing the steps right so I posted on here. I think the answer is 20...I don't know what else it could be :(

Answer 8

oh great! you've tried some steps...
can you post them here, we'll correct then if required.

Answer 9

Oh boy... I don't know where to begin! I'll write the most recent steps I've tried...one moment

Answer 10

I used the formula \[a=\log _{b}(b ^{a})\]

Answer 11

\[2= \log _{10}(10^2)\]

Answer 12

for the right side?

so, \(2 = \log 10^2 \)

Answer 13

but for the left side we will need the formula I posted :)

Answer 14

\(\log x + \log(x-15) = \log x\times(x-15) \\= \log 10^2\)

Answer 15

\[\log _{10}(x)+\log _{10}(x-15)=\log _{10}(10^2)\]

Answer 16

good

Answer 17

\[\log _{10}(x(x-15))=\log _{10}(10^2)\]

Answer 18

you get how
\(\Large \log x + \log (x-15) = \log [x\times (x-15)]\)

basically just a log property

Answer 19

and the answeres are x=20 and x=-5 but we cant have a negative log so it must be 20

Answer 20

if log A = log B,
then A = B

what that says is that you can just cancel out log on both sides!

so we have

x(x-15) = 10^2

got this?

Answer 21

yes I just skipped the last steps at the end because they were really small and by then I already figured out my answers :)

Answer 22

oh you solved it completely.
excellent! x =20 is correct :)

Answer 23

Thanks @hartnn , I appreciate your time :)

Answer 24

welcome ^_^
happy to help :)