How do you solve $log (x + 2) + log (x - 1) = 1$ $Log(x+2)+log(x-1)=1$ ?

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Answer 1 · 2016-02-29T01:38:28

Use the log rule ${log}_{a} m + {log}_{a} n = {log}_{a} (m \times n)$ $log_am + log_an = log_a(m xx n)$

$log (x + 2) + log (x - 1) = 1$ $log(x + 2) + log(x - 1) = 1$

$log (x + 2) (x - 1) = 1$ $log(x + 2)(x - 1)= 1$

$log (x^{2} + 2 x - x - 2) = 1$ $log(x^2 + 2x - x - 2) = 1$

$log (x^{2} + x - 2) = 1$ $log(x^2 + x - 2) = 1$

Convert to exponential form.

$x^{2} + x - 2 = 10^{1}$ $x^2 + x - 2 = 10^1$

$x^{2} + x - 12 = 0$ $x^2 + x - 12 = 0$

$(x + 4) (x - 3) = 0$ $(x + 4)(x - 3) = 0$

$x = - 4 and 3$ $x = -4 and 3$

Since the log of a negative number is undefined the solution is x = 3.

Hopefully this helps.

How do you solve log(x+2)+log(x−1)=1Log(x+2)+log(x-1)=1 ?