Math Doubts

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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