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

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Answer 1 · 2016-02-29T01:38:28

Use the log rule $log_am + log_an = log_a(m xx n)$

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

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

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

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

Convert to exponential form.

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

$x^2 + x - 12 = 0$

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

$x = -4 and 3$

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)=1 Log(x+2)+log(x-1)=1 ?