Question

log x + log(x-3)=1
help!

Answer 1

\[\log x(x-3)=1\]

Answer 2

\[\log a + \log b = \log ab\]\[\log a - \log b = \log a/b\]

Answer 3

simplify the logs 1st

\[\log(a) + \log(b) = \log(ab)\]

you need to simplify the left hand side using the law.

next raise both sides to the power of 10. Assuming that its a base 10 log.


the law says

\[10^{\log_{10} (a)} = a\]

and the right hand side will be 10^1

now solve for x

Answer 4

\[10^1=x(x-3)\]\[10=x^2-3x\]\[0=x^2-3x-10\]\[0=(x-5)(x+2)\]\[x=5,-2\]

Answer 5

log [x(x - 3)] = 1
log(x^2 - 3x) = 1
Using def of log:
x^2 - 3x = 10^1
x^2 - 3x - 10 = 0
(x - 5)(x + 2) = 0
x = 5 or x = -2
Since log(-2) is not defined,
x = 5

Answer 6

\[\log_{10} x=y \rightarrow 10^y=x\]

Answer 7

you need to ignore -2 as a solution as you can't take the log of a negative number.

Answer 8

Good point

Answer 9

So basically when you get rid of the logs from one side, the other side becomes 10?

Answer 10

yes... or the base of the logs

is it was base e..... e.g ln(x) = 5

x = e^5

Answer 11

ok got it thanks!

Answer 12

if not written otherwise. log is always base 10. :) Good luck

Answer 13

thank you!