Solve for x:
2log(x)-log(x=1)= log(4)-log(3)
OMG, I love solving logs!!
Are you familiar with the log laws Soulwraith? If not, I strongly advise you thoroughly revise them!
As for your question:
2*log(x)-log(x+1)= log(4)-log(3)
Here is a process to work it out (Assuming log is to the base 10):
$\displaystyle 2*\log x-\log (x+1)=\log 4-\log 3$
Simplifying the logs into single terms using log laws:
$\displaystyle log x^2-\log (x+1)=\log (4/3)$
$\displaystyle log (x^2/x+1)=\log (4/3)$
Taking the log of both sides leaves us with an equation we can solve for x:
$\displaystyle (x^2)/(x+1)=\frac{4}{3}$
Can you take it from here?
2log(x)-log(x+1)= log(4)-log(3), if this correct, we may continue
rule of logarithm
i) log xy = log x + log y
ii) log x/y = log x - log y
iii) log x^n = n log x
2log(x)-log(x+1)= log(4)-log(3),
use rule iii)
log x^2 - log (x + 1) = log 4 - log 3
use rule ii)
log ((x^2)/(x + 1) = log 4/3
take anti-log of both sides
(x^2)/(x + 1) = 4/3
cross-multiply
3x^2 = 4(x + 1),
3x^2 = 4x + 4,
3x^2 - 4x - 4 = 0,
factoring, we have
(x - 2)(3x + 2) = 0
x = {-2/3, 2}
but only x = 2 is valid.
for x to have the value of -2/3 is multivalued in logarithm.
the graph of log x^2 - log (x + 1) = log 4 - log 3 is shown below.
It confirms that x = 2 is only the valid value of x.