# How do you solve? log(x-1) + log 2x= log2x^2 - log2

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Nimo N. Share
Feb 19, 2018

The solution is:  color(brown)( x = 2

#### Explanation:

How do you solve
 color(blue)( log (x - 1) + log (2x) = log (2x^2) - log (2)

• Please note: It is assumed that an expression like log(x) means take the common logarithm, i.e. base 10 logarithm of x. If another base is intended, it should be indicated, as in  color(brown)( log_2 (x) . or use  color(brown)( ln (x)  for the "natural, base e logarithm".

Using properties of logarithms, we can combine terms, then rearrange terms to isolate x, if possible. Also, we must assume  color(blue)( 1 < x , otherwise the equation would not be valid for a real valued logarithm function.

$\log \left(x - 1\right) + \log \left(2 x\right) = \log \left(2 {x}^{2}\right) - \log \left(2\right)$
$\log \left(\left(x - 1\right) \cdot \left(2 x\right)\right) = \log \left(\frac{2 {x}^{2}}{2}\right)$

If two logarithm expressions with the same base are equal then the arguments of the functions are equal.
$\left(x - 1\right) \cdot \left(2 x\right) = \left(\frac{2 {x}^{2}}{2}\right)$
Simplify and solve for x.
$2 {x}^{2} - 2 x = {x}^{2}$
Subtract ${x}^{2}$ from both sides.
${x}^{2} - 2 x = 0$
$x \cdot \left(x - 2\right) = 0$

Either $x = 0$, which is not allowed,
or
$\left(x - 2\right) = 0$
 color(brown)( x = 2

We're not finished until we check, since there may have been an error in the work, or the only possible solution may not work.
 log ((2) - 1) + log (2(2)) =^? log (2(2)^2) - log (2)
 log (1) + log (4) =^? log (8) - log (2)
 0 + log (4) =^? log (8/2)
 log (4) =^? log (4) , YES!!! We have our solution.

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