# Can anybody please solve it.question is related to logarithms, range?

Apr 27, 2018

$0 \le x \le 1$ and $x = 4$ are all solutions to this equation.

#### Explanation:

First note that by using the change of base formula

${\log}_{9} x = \frac{{\log}_{3} x}{{\log}_{3} \text{9}} = \frac{{\log}_{3} x}{2}$.

Therefore

$2 {\log}_{9} x = {\log}_{3} x$, and

${\log}_{3} x = {\log}_{9} {x}^{2}$.

So we can transform

${\log}_{3} \left(\sqrt{x} + \left\mid \sqrt{x} - 1 \right\mid\right) = {\log}_{9} \left(4 \sqrt{x} - 3 + 4 \left\mid \sqrt{x} - 1 \right\mid\right)$

to

${\log}_{9} {\left(\sqrt{x} + \left\mid \sqrt{x} - 1 \right\mid\right)}^{2} = {\log}_{9} \left(4 \sqrt{x} - 3 + 4 \left\mid \sqrt{x} - 1 \right\mid\right)$.

Since the logarithm is a function, if the logs are equal then their arguments are equal.

${\left(\sqrt{x} + \left\mid \sqrt{x} - 1 \right\mid\right)}^{2} = 4 \sqrt{x} - 3 + 4 \left\mid \sqrt{x} - 1 \right\mid$

Because of the absolute value expressions, there are two cases here - when $x \ge 1$ and when $0 \le x < 1$.

When $x \ge 1$ we can ignore the absolute value signs and write

${\left(2 \sqrt{x} - 1\right)}^{2} = 4 \sqrt{x} - 3 + 4 \left(\sqrt{x} - 1\right)$

$4 x - 4 \sqrt{x} + 1 = 8 \sqrt{x} - 7$

$4 x - 12 \sqrt{x} + 8 = 0$

$x - 3 \sqrt{x} + 2 = 0$

$\left(\sqrt{x} - 2\right) \left(\sqrt{x} - 1\right) = 0$

$\sqrt{x} = 2$ and $\sqrt{x} = 1$ which means that $x = 4$, and $x = 1$ are both solutions. In fact, if we plug these values for $x$ back into the original equation, we can see that both of these values work.

For $x = 1$

${\log}_{3} \left(\sqrt{1} + \left\mid \sqrt{1} - 1 \right\mid\right) = {\log}_{9} \left(4 \sqrt{1} - 3 + 4 \left\mid \sqrt{1} - 1 \right\mid\right)$

${\log}_{3} \left(1\right) = {\log}_{9} \left(1\right)$

$0 = 0$.

For $x = 4$

${\log}_{3} \left(\sqrt{4} + \left\mid \sqrt{4} - 1 \right\mid\right) = {\log}_{9} \left(4 \sqrt{4} - 3 + 4 \left\mid \sqrt{4} - 1 \right\mid\right)$

${\log}_{3} \left(3\right) = {\log}_{9} \left(9\right)$

$1 = 1$.

Now for when $0 \le x < 1$, we can rewrite the equation as

(sqrt(x)-sqrt(x)+1))^2=4sqrt(x)-3+4(-sqrt(x)+1)

$1 = 1$

Because this is ALWAYS true, there are an infinite number of solutions between and including 0 and 1.