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Solve $\dfrac{\log{(\sqrt{x+1}+1)}}{\log{\sqrt[\Large 3]{x-40}}} \,=\, 3$

The quotient of the common logarithm of square root of $x$ plus $1$ plus $1$ by the logarithm of cube root of $x$ minus $40$ to base $10$ is equal to $3$ in this logarithmic problem.

$\dfrac{\log{\big(\sqrt{x+1}+1\big)}}{\log{\sqrt[\Large 3]{x-40}}}$ $\,=\,$ $3$

Now, let us learn how to solve this common logarithmic equation to find the value of $x$.

Simplify the Logarithmic equation

The cube root inside the logarithmic function in the denominator can be expressed in exponential notation.

$\implies$ $\dfrac{\log{\big(\sqrt{x+1}+1\big)}}{\log{(x-40)^{\Large \frac{1}{3}}}}$ $\,=\,$ $3$

Use the power rule of logarithms to write the exponent inside the function as a multiplying factor.

$\implies$ $\dfrac{\log{\big(\sqrt{x+1}+1\big)}}{\dfrac{1}{3} \times \log{(x-40)}}$ $\,=\,$ $3$

Now, continue the procedure of simplifying the logarithmic equation.

$\implies$ $\dfrac{\log{\big(\sqrt{x+1}+1\big)}}{\log{(x-40)}}$ $\,=\,$ $\dfrac{1}{3} \times 3$

$\implies$ $\dfrac{\log{\big(\sqrt{x+1}+1\big)}}{\log{(x-40)}}$ $\,=\,$ $\dfrac{1 \times 3}{3}$

$\implies$ $\dfrac{\log{\big(\sqrt{x+1}+1\big)}}{\log{(x-40)}}$ $\,=\,$ $\dfrac{3}{3}$

$\implies$ $\dfrac{\log{\big(\sqrt{x+1}+1\big)}}{\log{(x-40)}}$ $\,=\,$ $\dfrac{\cancel{3}}{\cancel{3}}$

$\implies$ $\dfrac{\log{\big(\sqrt{x+1}+1\big)}}{\log{(x-40)}}$ $\,=\,$ $1$

$\implies$ $\log{\big(\sqrt{x+1}+1\big)}$ $\,=\,$ $1 \times \log{(x-40)}$

$\implies$ $\log{\big(\sqrt{x+1}+1\big)}$ $\,=\,$ $\log{(x-40)}$

The equation expresses that the logarithmic expressions are equal. Hence, the functions inside the logarithms are equal.

$\,\,\,\therefore\,\,\,\,\,\,$ $\sqrt{x+1}+1$ $\,=\,$ $x-40$

Simplify the Algebraic equation

The logarithmic equation is successfully simplified as an algebraic equation. Now, let us focus on simplifying the algebraic equation for solving the value of the variable $x$.

$\implies$ $\sqrt{x+1}+1$ $\,=\,$ $x-40$

$\implies$ $\sqrt{x+1}$ $\,=\,$ $x-40-1$

$\implies$ $\sqrt{x+1}$ $\,=\,$ $x-41$

For eliminating the square root on left-hand side of the equation, take square on both sides of the equation.

$\big(\sqrt{x+1}\big)^2$ $\,=\,$ $(x-41)^2$

Expand the right hand side square of the subtraction of two terms.

$\implies$ $x+1$ $\,=\,$ $x^2+{41}^2-2 \times x \times 41$

$\implies$ $x+1$ $\,=\,$ $x^2+1681-82x$

$\implies$ $x+1$ $\,=\,$ $x^2-82x+1681$

$\implies$ $0$ $\,=\,$ $x^2-82x+1681-x-1 $

$\implies$ $x^2-82x-x+1681-1$ $\,=\,$ $0$

$\,\,\,\therefore\,\,\,\,\,\,$ $x^2-83x+1680$ $\,=\,$ $0$

Solve the Quadratic equation

The quadratic equation $x^2-83x+1680$ $\,=\,$ $0$ can be solved by using quadratic formula method.

$\implies$ $x$ $\,=\,$ $\dfrac{-(-83) \pm \sqrt{{(-83)}^2 -4 \times 1 \times 1680}}{2 \times 1}$

$\implies$ $x$ $\,=\,$ $\dfrac{83 \pm \sqrt{6889 -6720}}{2}$

$\implies$ $x$ $\,=\,$ $\dfrac{83 \pm \sqrt{169}}{2}$

$\implies$ $x$ $\,=\,$ $\dfrac{83 \pm 13}{2}$

$\implies$ $x$ $\,=\,$ $\dfrac{83+13}{2}$ or $x = \dfrac{83-13}{2}$

$\implies$ $x$ $\,=\,$ $\dfrac{96}{2}$ or $x = \dfrac{70}{2}$

$\,\,\,\therefore\,\,\,\,\,\,$ $x$ $\,=\,$ $48$ or $x = 35$

Therefore, the solution set of the quadratic equation $x^2-83x+1680$ $\,=\,$ $0$ is $\big\{35, 48\big\}$


Let us validate the both zeros by substituting each root in the given logarithmic equation.

Substitute $x \,=\, 35$ in the expression on left-hand side of the equation and find its value.

$=\,\,\,$ $\dfrac{\log{\big(\sqrt{35+1}+1\big)}}{\log{\sqrt[\Large 3]{35-40}}}$

$=\,\,\,$ $\dfrac{\log{\big(\sqrt{36}+1\big)}}{\log{\sqrt[\Large 3]{-5}}}$

Look at the denominator, the cube root of a negative number $-5$ is indeterminate. Hence, the value of the above logarithmic expression cannot be equal to $3$ when $x$ is equal to $35$.

$\therefore\,\,\,$ $\dfrac{\log{\big(\sqrt{36}+1\big)}}{\log{\sqrt[\Large 3]{-5}}}$ $\,\ne\,$ $3$

Therefore, the $x \,=\, 35$ cannot be a zero or root of the given logarithmic equation.

Now, substitute $x \,=\, 48$ in left-hand side expression and find its value.

$=\,\,\,$ $\dfrac{\log{\big(\sqrt{48+1}+1\big)}}{\log{\sqrt[\Large 3]{48-40}}}$

$=\,\,\,$ $\dfrac{\log{\big(\sqrt{49}+1\big)}}{\log{\sqrt[\Large 3]{8}}}$

$=\,\,\,$ $\dfrac{\log{(7+1)}}{\log{\sqrt[\Large 3]{2^3}}}$

$=\,\,\,$ $\dfrac{\log{8}}{\log{2}}$

$=\,\,\,$ $\dfrac{\log{(2^3)}}{\log{2}}$

$=\,\,\,$ $\dfrac{3 \times \log{2}}{\log{2}}$

$=\,\,\,$ $3 \times \dfrac{\log{2}}{\log{2}}$

$=\,\,\,$ $3 \times \dfrac{\cancel{\log{2}}}{\cancel{\log{2}}}$

$=\,\,\,$ $3 \times 1$

$=\,\,\,$ $3$

Therefore, the value of the logarithmic expression $\dfrac{\log{(\sqrt{x+1}+1)}}{\log{\sqrt[\Large 3]{x-40}}}$ is equal to $3$ when the value of $x$ is equal to $48$. Hence, $x \,=\, 48$ is the root or zero of the given common logarithmic equation.

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