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

Nov 18, 2015

$x = - 4 \text{ or } + 3$

Explanation:

Let "log base" be b

Known that ${\log}_{b} \left(b\right) = 1$

Assumption: The use of 'log' in the question is referring to logs to base 10 giving:

${\log}_{10} \left(x + 2\right) + {\log}_{10} \left(x - 1\right) = 1. \ldots \ldots \ldots \ldots \ldots \ldots . . \left(1\right)$

But $\textcolor{w h i t e}{\times \times \times} {\log}_{10} \left(10\right) = 1 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(2\right)$

Substitute (2) into (1) giving:

${\log}_{10} \left(x + 2\right) + {\log}_{10} \left(x - 1\right) = {\log}_{10} \left(10\right) \ldots \ldots \ldots . \left(3\right)$

But ${\log}_{10} \left(x + 2\right) + {\log}_{10} \left(x - 1\right) = {\log}_{10} \left[\left(x + 2\right) \left(x - 1\right)\right]$

Rewrite (3) as:

${\log}_{10} \left[\left(x + 2\right) \left(x - 1\right)\right] = {\log}_{10} \left(10\right)$
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Personal note: I am of 'very' old school where $\text{antilog} \to {\log}^{- 1}$

Taking antilogs $\to \left(x + 2\right) \left(x - 1\right) = 10$

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

$\left(x + 4\right) \left(x - 3\right) = 0$

$\textcolor{b l u e}{x = - 4 \text{ or } + 3}$