Question

How do you solve $\dfrac{\log \left( 35-{{x}^{3}} \right)}{\log \left( 5-x \right)}=3$?

Answer 1

Hint: Use the formula $a\log x=\log {{x}^{a}}$ to convert the left side and right side of the equation in terms of the logarithm function of $x$. Take antilogarithm from both sides and solve the polynomial form in terms of $x$. Use the expansion formula ${{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}+3a{{b}^{2}}-3{{a}^{2}}b$ and then solve the obtained polynomial to find the value of $x$.

Complete step by step solution:
Multiply both sides of the polynomial with $\log \left( 5-x \right)$.
$\Rightarrow \dfrac{\log \,\left( 35-{{x}^{3}} \right)}{\log \,\left( 5-x\, \right)}\times \log \left( 5-x \right)=3\times \log \left( 5-x \right)$
Now, when we cancel log(5-x), we get,
$\Rightarrow \log \,\left( 35-{{x}^{3}} \right)=3\log \,\left( 5-x \right)$
Apply the formula $a\log x=\log {{x}^{a}}$ to rewrite the equation.
$\Rightarrow \log \left( 35-{{x}^{3}} \right)=\log {{\left( 5-x \right)}^{3}}$
Take antilog of both sides of the equation.
$\Rightarrow {{e}^{\log \left( 35-{{x}^{3}} \right)}}={{e}^{\log {{\left( 5-x \right)}^{3}}}}$
When we simplify the above equation, we get,
$\Rightarrow 35-{{x}^{3}}={{\left( 5-x \right)}^{3}}$
Use the expansion formula ${{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}+3a{{b}^{2}}-3{{a}^{2}}b$ to expand the right side of the equation.
$\Rightarrow {{\left( 5-x \right)}^{3}}={{\left( 5 \right)}^{3}}-{{\left( x \right)}^{3}}+3\left( 5 \right){{\left( x \right)}^{2}}-3{{\left( 5 \right)}^{2}}\left( x \right)$
$\Rightarrow {{\left( 5-x \right)}^{3}}=125-{{x}^{3}}+15{{x}^{2}}-75x$
Substitute the value of ${{\left( 5-x \right)}^{3}}$ in the formula.
$\Rightarrow 35-{{x}^{3}}=125-{{x}^{3}}+15{{x}^{2}}-75x$
Add ${{x}^{3}}$ on both sides of the equation.
$\Rightarrow 35-{{x}^{3}}+{{x}^{3}}-35=125-{{x}^{3}}+15{{x}^{2}}-75x+{{x}^{3}}-35$
After simplifying the above equation, we get,
$\Rightarrow 15{{x}^{2}}-75x+90=0$
Divide the whole equation by 15.
$\Rightarrow \dfrac{15{{x}^{2}}}{15}-\dfrac{75x}{15}+\dfrac{90}{15}=\dfrac{0}{15}$
After simplifying the above equation, we get,
$\Rightarrow {{x}^{2}}-5x+6=0$
Factorize the obtained quadratic equation by splitting the middle term $5x$ into $-3x$ and $-2x$.
$\Rightarrow {{x}^{2}}-3x-2x+6=0$
After simplifying the above equation, we get,
$\Rightarrow x\left( x-3 \right)-2\left( x-3 \right)=0$
Factor out $\left( x-3 \right)$ from both the terms.
$\Rightarrow \left( x-3 \right)\left( x-2 \right)=0$
Equate factor $\left( x-3 \right)$ with zero to find the value of $x$.
$\Rightarrow x-3=0$
After simplifying the above equation, we get,
$\Rightarrow x=3$
Equate factor $\left( x-2 \right)$ with zero to find the value of $x$.
$\Rightarrow x-2=0$

$\Rightarrow x=2$

Note:
Whenever we have to solve the equation for some variable $x$ containing terms of logarithm and non-logarithmic terms then we try to convert each side of the equation in terms of logarithm so that it can be solved by taking logarithm of both sides. Use of properties like $a\log x=\log {{x}^{a}}$, $\log x + \log y=\log xy$ helps to obtain such configuration.