How do you solve $2 {log}_{5} x = {log}_{5} 9$ $2log_5x = log_5 9$ ?

Question

3762 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-24T01:37:22

First simplify using the rule $a log n = {log n}^{a}$ $alogn = logn^a$

${log}_{5} (x^{2}) = {log}_{5} (9)$ $log_5(x^2) = log_5(9)$

Put everything to one side of the equation.

$0 = {log}_{5} (9) - {log}_{5} (x^{2})$ $0 = log_5(9) - log_5(x^2)$

Simplify using the rule ${log}_{a} m - {log}_{a} n = {log}_{a} (\frac{m}{n})$ $log_am - log_an = log_a(m / n)$

$0 = {log}_{5} (\frac{9}{x^{2}})$ $0 = log_5(9/x^2)$

Convert to exponential form.

$5^{0} = \frac{9}{x^{2}}$ $5^0 = 9/x^2$

$1 = \frac{9}{x^{2}}$ $1 = 9/x^2$

$x^{2} = 9$ $x^2 = 9$

$x = \pm 3$ $x = +-3$

However, only +3 works as a solution because the log of a negative number is non-defined.

The solution set is x = 3.

Hopefully this helps!

How do you solve 2log5x=log592log_5x = log_5 9?