Find the value of $x$ if $x^{(\log_{2} x) + 4} = 32$

$x$ is a variable and the sum of logarithm of $x$ to base $2$ and number $4$ become an exponent to the literal $x$. The entire functional value is equal to $32$. The exponential logarithmic equation can be solved to find the value of $x$.

Step: 1

Express this exponential form equation in logarithmic form by using the fundamental relation between logarithmic and exponential functions.

$\implies$ $\large \log_{2} x$ $+ 4 = \large \log_{x} \normalsize 32$

$\implies 4 = \large \log_{x} \normalsize 32 -\large \log_{2} x$

$\implies \large \log_{x} \normalsize 32 -\large \log_{2} x \normalsize = 4$

$\implies \large \log_{x} \normalsize 2^5 -\large \log_{2} x \normalsize = 4$

Step: 2

Use the power rule of logarithms to write the first logarithmic term in simple form.

$\implies 5 \times \large \log_{x} \normalsize 2 -\large \log_{2} x \normalsize = 4$

$\implies 5 \large \log_{x} \normalsize 2 -\large \log_{2} x \normalsize = 4$

Remember this logarithmic equation and it is solved in two different methods to get the value of $x$. Use change of base rule of logarithms to transform either $\large \log_{x} \normalsize 2$ as $\large \log_{2} x$ or vice-versa.

Method: 1

$\implies 5 \large \log_{x} \normalsize 2 -\large \log_{2} x \normalsize = 4$

In this method, express $\large \log_{x} \normalsize 2$ in its reciprocal form by using base changing rule of logarithms.

$\implies 5 \times \large \dfrac{1}{\log_{2} x} -\log_{2} x \normalsize = 4$

$\implies \large \dfrac{\normalsize 5}{\log_{2} x} -\log_{2} x \normalsize = 4$

Step: 3

Take $\large \log_{2} x \normalsize = y$ and transform this logarithmic equation in terms of $y$.

$\implies \dfrac{5}{y} -y = 4$

$\implies \dfrac{5-y^2}{y} = 4$

$\implies 5- y^2= 4y$

$\implies 5= y^2 + 4y$

$\implies y^2 + 4y = 5$

$\implies y^2 + 4y -5 = 0$

Step: 4

It is a quadratic equation and it can be solved by the factoring method.

$\implies y^2 + 5y -y -5 \times 1 = 0$

$\implies y(y+ 5)-1(y+5) = 0$

$\implies (y-1)(y+5) = 0$

$\therefore y = 1$ and $y = -5$

Step: 5

But the variable $y = \large \log_{2} x$ as per our assumption in this method.

Therefore, $\large \log_{2} x \normalsize = 1$ and $\large \log_{2} x \normalsize = -5$

Now, express, each logarithmic function in exponential notation.

$\implies x = 2^1$ and $x = 2^{-5}$

$\implies x = 2$ and $x = \dfrac{1}{2^5}$

$\therefore \,\,\,\,\,\, x = 2$ and $x = \dfrac{1}{32}$

Therefore, it is the required solution of this problem. The literal $x$ has two values and they are $2$ and $\dfrac{1}{32}$.

Method: 2

Come back to our logarithmic equation, $5 \large \log_{x} \normalsize 2 -\large \log_{2} x \normalsize = 4$

Now, express $\large \log_{2} \normalsize x$ in its reciprocal form by using base changing rule of logarithms.

$\implies 5 \large \log_{x} \normalsize 2 -\dfrac{1}{\large \log_{x} \normalsize 2} \normalsize = 4$

Step: 3

Now, assume $\large \log_{x} \normalsize 2 = z$.

$\implies 5z -\dfrac{1}{z} = 4$

$\implies \dfrac{5z^2-1}{z} = 4$

$\implies 5z^2-1 = 4z$

$\implies 5z^2-4z-1 = 0$

Step: 4

It is a quadratic equation and solve it by quadratic formula method.

$z = \dfrac{-(-4) \pm \sqrt{{(-4)}^2 -4 \times 5 \times (-1)}}{2 \times 5}$

$\implies z = \dfrac{4 \pm \sqrt{16+20}}{10}$

$\implies z = \dfrac{4 \pm \sqrt{36}}{10}$

$\implies z = \dfrac{4 \pm 6}{10}$

$\implies z = \dfrac{4+6}{10}$ and $z = \dfrac{4-6}{10}$

$\implies z = \dfrac{10}{10}$ and $z = \dfrac{-2}{10}$

$\require{cancel} \implies z = \dfrac{\cancel{10}}{\cancel{10}}$ and $z = \dfrac{\cancel{-2}}{\cancel{10}}$

$\therefore \,\,\,\,\,\, z = 1$ and $z = \dfrac{-1}{5}$

Step: 5

Now replace $z$ by its actual value and solve the each solution to obtain the answer for this problem.

$\implies \large \log_{x} \normalsize 2 = 1$ and $\large \log_{x} \normalsize 2 = \dfrac{-1}{5}$

$\implies 2 = x^1$ and $2 = x^{\frac{-1}{5}}$

$\implies 2 = x$ and $2 = \dfrac{1}{x^{\frac{1}{5}}}$

$\implies x = 2$ and $x^{\frac{1}{5}} = \dfrac{1}{2}$

$\implies x = 2$ and $x = {\Big(\dfrac{1}{2}\Big)}^5$

$\implies x = 2$ and $x = \dfrac{1}{2^5}$

$\therefore \,\,\,\,\,\, x = 2$ and $x = \dfrac{1}{32}$

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