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

$x$ is a literal and involved in a mathematical expression through surd and logarithmic form to represent the quantity $3$. The value of $x$ is unknown and we have to find the value to know for which value, the given function equals to number $3$.

The value of cube root is $\dfrac{1}{3}$. So, express the cube root in denominator as an exponent $\dfrac{1}{3}$ for the function.

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

Use power law of logarithm to write the exponent of the function as a multiple of logarithmic function in denominator.

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

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

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

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

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

Use cross multiplication rule to try to express the equation in a simplified manner.

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

The logarithms of both sides is common logarithm and it states that they are equal mathematically. So, the functions should be equal.

Solve the equation to obtain the value of $x$

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

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

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

Take square both sides.

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

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 $

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

$\implies x^2-83x+1680 = 0$

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

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

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

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

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

$\implies x = \dfrac{96}{2}$ and $x = \dfrac{70}{2}$

$\implies x = 48$ and $x = 35$

$\implies x = 48$ and $35$

The quadratic equation is given two solutions but in some cases only one value is true value and other is false. So, let us try to find true root or roots of this equation.

Substitute $x = 48$ in left hand side of the given logarithmic equation.

$\implies \dfrac{\log(\sqrt{x+1}+1)}{\log \sqrt[3]{x-40}}$ $=$ $\dfrac{\log(\sqrt{48+1}+1)}{\log \sqrt[3]{48-40}}$

$\implies \dfrac{\log(\sqrt{x+1}+1)}{\log \sqrt[3]{x-40}}$ $=$ $\dfrac{\log(\sqrt{49}+1)}{\log \sqrt[3]{48-40}}$

$\implies \dfrac{\log(\sqrt{x+1}+1)}{\log \sqrt[3]{x-40}}$ $=$ $\dfrac{\log(7+1)}{\log \sqrt[3]{8}}$

$\implies \dfrac{\log(\sqrt{x+1}+1)}{\log \sqrt[3]{x-40}}$ $=$ $\dfrac{\log(8)}{\log \sqrt[3]{2^3}}$

$\implies \dfrac{\log(\sqrt{x+1}+1)}{\log \sqrt[3]{x-40}}$ $=$ $\dfrac{\log(2^3)}{\log 2}$

$\implies \dfrac{\log(\sqrt{x+1}+1)}{\log \sqrt[3]{x-40}}$ $=$ $\dfrac{3 \log 2}{\log 2}$

$\implies \dfrac{\log(\sqrt{x+1}+1)}{\log \sqrt[3]{x-40}}$ $=$ $3 \times \dfrac{\log 2}{\log 2}$

$\implies \require{cancel} \dfrac{\log(\sqrt{x+1}+1)}{\log \sqrt[3]{x-40}}$ $=$ $3 \times \dfrac{\cancel{\log 2}}{\cancel{\log 2}}$

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

It is verified that $x = 48$ is true solution for this problem and check $x = 35$.

$\implies \dfrac{\log(\sqrt{x+1}+1)}{\log \sqrt[3]{x-40}}$ $=$ $\dfrac{\log(\sqrt{35+1}+1)}{\log \sqrt[3]{35-40}}$

$\implies \dfrac{\log(\sqrt{x+1}+1)}{\log \sqrt[3]{x-40}}$ $=$ $\dfrac{\log(\sqrt{36}+1)}{\log \sqrt[3]{-5}}$

The cube root is undefined for negative numbers. So, the value of $x$ is not equal to $35$. In other words, the value of $x$ is $48$ and it is the required solution for this problem.

Latest Math Topics

Latest Math Problems

Email subscription

Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers.
Know more

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.