Evaluate log base 2 of 4x-3 = log base 2 of 3+ log base 2 of x

$\log_{2} (4 x) - 3 = \log_{2} (3) + \log_{2} (x)$

Step 1

Simplify the right side.

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\log_{2} (4 x) - 3 = \log_{2} (3 x)$

Step 2

Move all the terms containing a logarithm to the left side of the equation.

$\log_{2} (4 x) - \log_{2} (3 x) = 3$

Step 3

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\log_{2} (\frac{4 x}{3 x}) = 3$

Step 4

Cancel the common factor of $x$ .

Cancel the common factor.

$\log_{2} (\frac{4 x}{3 x}) = 3$

Rewrite the expression.

$\log_{2} (\frac{4}{3}) = 3$

Step 5

The equation is never true.

No solution