How do you solve Log_2 (10X + 4) - Log_2 X = 3?

Jan 5, 2016

No solution.

Explanation:

First, simplify using the rule that ${\log}_{a} b - {\log}_{a} c = {\log}_{a} \left(\frac{b}{c}\right)$.

${\log}_{2} \left(\frac{10 X + 4}{X}\right) = 3$

To undo the logarithm, exponentiate both sides with base $2$.

${2}^{{\log}_{2} \left(\frac{10 X + 4}{X}\right)} = {2}^{3}$

$\frac{10 X + 4}{X} = 8$

Solve for $X$.

$10 X + 4 = 8 X$

$X = - 2$

Warning! This is an invalid answer. If $X = - 2$, then both of the logarithm functions in the original equation would have a negative argument. It's impossible to take the logarithm of a negative number.

If we graph this as a function, the graph should never cross the $x$-axis, indicating a lack of roots.

graph{ln(10x+4)/ln2-lnx/ln2-3 [-5.64, 22.84, -3.47, 10.77]}