1. ## Logs

Solve for x:
2log(x)-log(x=1)= log(4)-log(3)

2. Hi Soulwraith56

Do you mean : 2*log(x)-log(x+1)= log(4)-log(3) ?

3. OMG, I love solving logs!!

Are you familiar with the log laws Soulwraith? If not, I strongly advise you thoroughly revise them!

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?

4. 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.