Math Doubts

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.

Math Questions

The math problems with solutions to learn how to solve a problem.

Learn solutions

Math Worksheets

The math worksheets with answers for your practice with examples.

Practice now

Math Videos

The math videos tutorials with visual graphics to learn every concept.

Watch now

Subscribe us

Get the latest math updates from the Math Doubts by subscribing us.

Learn more

Math Doubts

A free math education service for students to learn every math concept easily, for teachers to teach mathematics understandably and for mathematicians to share their maths researching projects.

Copyright © 2012 - 2023 Math Doubts, All Rights Reserved