# If $\log_{x} 2 \times \log_{\frac{x}{16}} 2$ $=$ $\log_{\frac{x}{64}} 2$, find the value of $x$

$x$ is a literal number. The product of logarithm of $2$ to $x$ and logarithm of $2$ to quotient of $x$ by $16$ is equal to the logarithm of $2$ to quotient of $x$ by $64$. The value of $x$ has to be evaluated on the basis of this data.

The three logarithmic terms contain different bases. So, it is not possible to simply it easily. However, the number $2$ is a common number to the three logarithmic terms. Hence, express each logarithmic term in reciprocal form.

01

### Interchanging the numbers and bases in log terms

The base and number can be interchanged in every logarithmic term by using the change of base rule of logarithms.

$\implies$ $\dfrac{1}{\log_{2} x} \times \dfrac{1}{\log_{2} \Big(\dfrac{x}{16}\Big)}$ $\,=\,$ $\dfrac{1}{\log_{2} \Big(\dfrac{x}{64}\Big)}$

$\implies$ $\dfrac{1}{\log_{2} x \times \log_{2} \Big(\dfrac{x}{16}\Big)}$ $\,=\,$ $\dfrac{1}{\log_{2} \Big(\dfrac{x}{64}\Big)}$

$\implies$ $\log_{2} \Big(\dfrac{x}{64}\Big)$ $\,=\,$ $\log_{2} x \times \log_{2} \Big(\dfrac{x}{16}\Big)$

$\implies$ $\log_{2} x \times \log_{2} \Big(\dfrac{x}{16}\Big)$ $=$ $\log_{2} \Big(\dfrac{x}{64}\Big)$

02

### Simplification of the equation

There are two logarithmic terms, which contain quantities in quotient form. They can be expanded by using the quotient rule of logarithmics.

$\implies$ $\log_{2} x$ $\times$ $(\log_{2} x -\log_{2} 16)$ $\,=\,$ $\log_{2} x -\log_{2} 64$

$\implies$ $(\log_{2} x) \times (\log_{2} x)$ $-(\log_{2} 16) \times (\log_{2} x)$ $\,=\,$ $\log_{2} x -\log_{2} 64$

$\implies$ ${(\log_{2} x)}^2$ $-(\log_{2} 2^4) \times (\log_{2} x)$ $\,=\,$ $\log_{2} x -\log_{2} 2^6$

Apply the power rule of logarithms to simplify the logarithmic equation.

$\implies$ ${(\log_{2} x)}^2$ $-(4\log_{2} 2) \times (\log_{2} x)$ $\,=\,$ $\log_{2} x -6\log_{2} 2$

Use logarithm of base rule to simplify the equation further.

$\implies$ ${(\log_{2} x)}^2 -(4 \times 1) \times (\log_{2} x)$ $\,=\,$ $\log_{2} x -6 \times 1$

$\implies$ ${(\log_{2} x)}^2 -4 \times (\log_{2} x)$ $\,=\,$ $\log_{2} x -6$

$\implies$ ${(\log_{2} x)}^2 -4 \log_{2} x$ $\,=\,$ $\log_{2} x -6$

$\implies$ ${(\log_{2} x)}^2 -4 \log_{2} x -\log_{2} x +6$ $\,=\,$ $0$

$\implies$ ${(\log_{2} x)}^2 -5 \log_{2} x +6$ $\,=\, 0$

The equation in logarithmic form represents a quadratic equation. It can be simplified by the solution of the quadratic equation.

03

### Solving the Quadratic equation

Take $y \,=\, \log_{2} x$ and transform the equation in terms of $y$ from logarithmic form.

$\implies$ $y^2 -5y +6 = 0$

$\implies$ $y^2 -3y -2y +6 = 0$

The quadratic equation can be solved easily by the factoring method.

$\implies$ $y(y-3) -2(y-3) = 0$

$\implies$ $(y-3)(y-2) = 0$

$\implies$ $y-3 = 0$ and $y-2 = 0$

$\therefore \,\,\,\,\,\,$ $y = 3$ and $y = 2$

Therefore, the values of $y$ are $2$ and $3$.

04

### Solution

Now, replace the actual value of the literal number $y$ to find the value of $x$.

$\therefore \,\,\,\,\,\,$ $\log_{2} x = 2$ and $\log_{2} x = 3$

Now, write each logarithmic equation in exponential notation to obtain the value of $x$ mathematically.

$\implies$ $x = 2^2$ and $x = 2^3$

$\therefore \,\,\,\,\,\,$ $x = 4$ and $x = 8$

The logarithmic equation has given two values for the literal $x$ as solution.

05

### Verifying the values

Substitute $x = 4$ in both sides of the equation.

$\implies$ $\log_{x} 2 \times \log_{\frac{x}{16}} 2$ $\,=\,$ $\log_{4} 2 \times \log_{\frac{4}{16}} 2$

$\implies$ $\log_{x} 2 \times \log_{\frac{x}{16}} 2$ $\,=\,$ $\require{cancel} \log_{2^2} 2 \times \log_{\frac{\cancel{4}}{\cancel{16}}} 2$

$\implies$ $\log_{x} 2 \times \log_{\frac{x}{16}} 2$ $\,=\,$ $\dfrac{1}{2} \log_{2} 2 \times \log_{\frac{1}{4}} 2$

$\implies$ $\log_{x} 2 \times \log_{\frac{x}{16}} 2$ $\,=\,$ $\dfrac{1}{2} \times 1 \times \log_{4^{-1}} 2$

$\implies$ $\log_{x} 2 \times \log_{\frac{x}{16}} 2$ $\,=\,$ $\dfrac{1}{2} \times \log_{2^{-2}} 2$

$\implies$ $\log_{x} 2 \times \log_{\frac{x}{16}} 2$ $\,=\,$ $\dfrac{1}{2} \log_{2^{-2}} 2$

$\implies$ $\log_{x} 2 \times \log_{\frac{x}{16}} 2$ $\,=\,$ $\dfrac{1}{2} \times \dfrac{-1}{2} \log_{2} 2$

$\implies$ $\log_{x} 2 \times \log_{\frac{x}{16}} 2$ $\,=\,$ $-\dfrac{1}{4} \times 1$

$\implies$ $\log_{x} 2 \times \log_{\frac{x}{16}} 2$ $\,=\,$ $-\dfrac{1}{4}$

Similarly,

$\implies$ $\log_{\frac{x}{64}} 2$ $\,=\,$ $\log_{\frac{4}{64}} 2$

$\implies$ $\log_{\frac{x}{64}} 2$ $\,=\,$ $\require{cancel} \log_{\frac{\cancel{4}}{\cancel{64}}} 2$

$\implies$ $\log_{\frac{x}{64}} 2$ $\,=\,$ $\log_{\frac{1}{16}} 2$

$\implies$ $\log_{\frac{x}{64}} 2$ $\,=\,$ $\log_{2^{-4}} 2$

$\implies$ $\log_{\frac{x}{64}} 2$ $\,=\,$ $-\dfrac{1}{4} \log_{2} 2$

$\implies$ $\log_{\frac{x}{64}} 2$ $\,=\,$ $-\dfrac{1}{4} \times 1$

$\implies$ $\log_{\frac{x}{64}} 2$ $\,=\,$ $-\dfrac{1}{4}$

Therefore, the values of the both sides of the equation is equal. Hence, the value of $x = 4$ is a true value.

Now, check the value $x = 8$

$\implies$ $\log_{x} 2 \times \log_{\frac{x}{16}} 2$ $\,=\,$ $\log_{8} 2 \times \log_{\frac{8}{16}} 2$

$\implies$ $\log_{x} 2 \times \log_{\frac{x}{16}} 2$ $\,=\,$ $\require{cancel} \log_{2^3} 2 \times \log_{\frac{\cancel{8}}{\cancel{16}}} 2$

$\implies$ $\log_{x} 2 \times \log_{\frac{x}{16}} 2$ $\,=\,$ $\dfrac{1}{3} \log_{2} 2 \times \log_{\frac{1}{2}} 2$

$\implies$ $\log_{x} 2 \times \log_{\frac{x}{16}} 2$ $\,=\,$ $\dfrac{1}{3} \times 1 \times \log_{2^{-1}} 2$

$\implies$ $\log_{x} 2 \times \log_{\frac{x}{16}} 2$ $\,=\,$ $\dfrac{1}{3} \times \dfrac{1}{-1} \log_{2} 2$

$\implies$ $\log_{x} 2 \times \log_{\frac{x}{16}} 2$ $\,=\,$ $-\dfrac{1}{3} \times (-1) \times 1$

$\implies$ $\log_{x} 2 \times \log_{\frac{x}{16}} 2$ $\,=\,$ $-\dfrac{1}{3}$

In the same way,

$\implies$ $\log_{\frac{x}{64}} 2$ $\,=\,$ $\log_{\frac{8}{64}} 2$

$\implies$ $\log_{\frac{x}{64}} 2$ $\,=\,$ $\require{cancel} \log_{\frac{\cancel{8}}{\cancel{64}}} 2$

$\implies$ $\log_{\frac{x}{64}} 2$ $\,=\,$ $\log_{\frac{1}{8}} 2$

$\implies$ $\log_{\frac{x}{64}} 2$ $\,=\,$ $\log_{2^{-3}} 2$

$\implies$ $\log_{\frac{x}{64}} 2$ $\,=\,$ $-\dfrac{1}{3} \log_{2} 2$

$\implies$ $\log_{\frac{x}{64}} 2$ $\,=\,$ $-\dfrac{1}{3} \times 1$

$\implies$ $\log_{\frac{x}{64}} 2$ $\,=\,$ $-\dfrac{1}{3}$

Therefore, the values of both sides of the logarithmic equation are same. So, the value $x = 8$ is also true value.

Therefore, the literal $x$ has two solutions and they are $4$ and $8$ and it is the required solution for this logarithmic problem in mathematics.

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