`log_3(x) + log_3(x - 2) = 1` Solve for x or b

Thursday, June 9, 2016

$\displaystyle{\log}_{{3}}{\left({x}\right)}+{\log}_{{3}}{\left({x}-{2}\right)}={1}$ Solve for x or b

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

The logarithms at the left side have the same base. So express the left side with one logarithm only using the rule $\displaystyle{\log}_{{b}}{\left({M}\right)}+{\log}_{{b}}{\left({N}\right)}={\log}_{{b}}{\left({M}\cdot{N}\right.}$ ).

$\displaystyle{\log}_{{3}}{\left({x}\cdot{\left({x}-{2}\right)}\right)}={1}$

$\displaystyle{\log}_{{3}}{\left({{x}}^{{2}}-{2}{x}\right)}={1}$

Then, convert this to exponential form.

Take note that if a logarithmic equation is in the form

$\displaystyle{y}={\log}_{{b}}{\left({x}\right)}$

its equivalent exponential equation is

$\displaystyle{x}={{b}}^{{y}}$

So converting

$\displaystyle{\log}_{{3}}{\left({{x}}^{{2}}-{2}{x}\right)}={1}$

to exponential equation, it becomes:

$\displaystyle{{x}}^{{2}}-{2}{x}={{3}}^{{1}}$

$\displaystyle{{x}}^{{2}}-{2}{x}={3}$

Now the equation is in quadratic form. To solve it, one side should be zero.

$\displaystyle{{x}}^{{2}}-{2}{x}-{3}={0}$

Factor the left side.

$\displaystyle{\left({x}-{3}\right)}{\left({x}+{1}\right)}={0}$

Set each factor equal to zero. And isolate the x.

$\displaystyle{x}-{3}={0}$

$\displaystyle{x}={3}$

$\displaystyle{x}+{1}={0}$

$\displaystyle{x}=-{1}$

Now that the values of x are known, consider the condition in a logarithm. The argument of a logarithm should always be positive.

In the equation

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

the arguments are x and x - 2. So the values of these two should all be above zero.

$\displaystyle{x}\gt{0}$

$\displaystyle{x}-{2}\gt{0}$

Between the two values of x that we got, it is only x = 3 that satisfy this condition.

Therefore, the solution is $\displaystyle{x}={2}$ .

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