How do you solve ${log}_{3} (x + 4) - {log}_{3} x = 2$ $log_3(x+4)-log_3x=2$ ?

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Answer 1 · 2015-12-14T17:32:30

$x = \frac{1}{2}$ $x = 1/2$

The relevant law of logarithms that will help us here is the following:

${log}_{z} A - {log}_{z} B = {log}_{z} (\frac{A}{B})$ $log_z A - log_z B = log_z (A/B)$

where $A$ $A$ and $B$ $B$ can be any expression and $z$ $z$ is the base.

Applying this law to our equation will give us

${log}_{3} (\frac{x + 4}{x}) = 2$ $log_3 ((x+4)/x) = 2$

From here we can simply raise $3$ $3$ to both sides of the equation.

$3^{{log}_{3} (\frac{x + 4}{x})} = 3^{2}$ $3^(log_3 ((x+4)/x)) = 3^2$

On the left-hand side, the $3$ $3$ will cancel with the ${log}_{3}$ $log_3$ leaving us with

$\frac{x + 4}{x} = 9$ $(x+4)/x = 9$

Solving for $x$ $x$ yields $x = \frac{1}{2}$ $x = 1/2$ .

How do you solve log3(x+4)−log3x=2log_3(x+4)-log_3x=2?