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

Jan 9, 2016

$x = - 4$

Explanation:

The basic approach to solving log equations it to get them into simplest form and then put them back into exponential form.

In the case of this problem, the log expressions are not as simple as possible and you need to use your log rules to combine the two expressions. Because there is not base written with LOG the base is assumed to be 10.

$\log \left[\left(x + 2\right) \cdot \left(x - 1\right)\right] = 1$
$\log \left[{x}^{2} + x - 2\right] = 1$

put the equation back into exponential form:

${10}^{1} = {x}^{2} + x - 2$
$10 = {x}^{2} + x - 2$
${x}^{2} + x - 12 = 0$
$\left(x + 4\right) \left(x - 3\right) = 0$
$x = - 4 \mathmr{and} 3$

Issues of domain exist with log equation since the domain of $y = \log x$ is x>0, therefore it is crucial to test answers to ensure they work. In this case, $x = - 4$ makes each log expression undefined therefore the only answer is $x = 3$.

Jan 9, 2016

Assuming the base of the logarithm is 10, the answer is 3.

Explanation:

Using the fact that log X + log Y = log XY, one gets $\log \left(x + 2\right) + \log \left(x - 1\right) = \log \left(\left(x + 2\right) \left(x - 1\right)\right) = \log \left({x}^{2} + x - 2\right) = 1$. That means ${10}^{\log \left({x}^{2} + x - 2\right)} = {10}^{1}$. So, by the definition of a logarithm, one gets ${x}^{2} + x - 2 = 10$, or ${x}^{2} + x - 12 = 0$. By the quadratic equation, one gets x =( (-1 (+-) sqrt(1-4*1*(-12))) /2 ).
That is, x = $\frac{- 1 \left(\frac{+}{-}\right) \sqrt{49}}{2}$ or x = -4 or x = 3. Since x can't be -4, as the logarithm of a negative number is undefined, one gets x=3.