Question

How do I set this up? Please help:

log x + log(x − 3) = 1

Answer 1

log a + log b = log ab

Answer 2

First apply that property. What do you get?

Answer 3

(log x)(log (x-3))?

Answer 4

drop the log swing the base and pick up its exponent???

Answer 5

What?

Answer 6

I'm sorry. I dont understand

Answer 7

so would it be 10^x-3=1?

Answer 8

log a + log b = log ab
log x + log(x-3)=log (x^2-3x)=1

Answer 9

then do i change into exponential form?

Answer 10

@Mertsj is this right? @radar : 10=x^2-3x for the next step?

Answer 11

yes.. that's correct.

Answer 12

now all you gotta do is solve that quadratic... then check your solution(s) for any extraneous ones...

Answer 13

do you get +/- 13?

Answer 14

square root of 13 i mean

Answer 15

10 = x^2 - 3x
0 = x^2 - 3x - 10
0 = (x - 5)(x + 2)
x = 5 or x = -2

now check which one works...

Answer 16

So i can also solve by a^2 + b^2 = c^2? so its 5

Answer 17

Thank you so much!!

Answer 18

\[\log_{10}{x}+ \log_{10}{(x-3)} =1.\]\[\implies \log_{10}{(x)(x-3)} =1\]Becoz\[\log_a{M}+\log_a{N} =\log_a{MN} . \]therefore\[x(x-3)=10^1.\]\[x^2-3x-10=0\]\[\implies x^2-5x+2x-10=0\]Now do it on ur ow☺

Answer 19

own*

Answer 20

I did 100=x^2 +3x^2 then solved and got 5

Answer 21

what r u doing!
solve like me in my sol.

Answer 22

Ok i got 5 and -2

Answer 23

ya u r correct.
but only one solution will be acceptable which is 5.
-2 will not be acceptable becoz In\[\log_{10}{(x-3)}\]If u put x=-2
then (x-3)<0
which say that log does not exist.
So as a summary only 5 will be answer