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# How to solve ${\log _x}36 + {\log _{18}}3x = 3$ ?

Last updated date: 20th Jun 2024
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Hint: Here, we will use the logarithmic rule to rewrite the given logarithmic equation. We will simplify the equation using different logarithmic rules to form a quadratic equation. Then by using we will find two values of the variable of the obtained quadratic equation. Then by solving the values of the variable we will get the required solution.

Formula Used:
We will use the following formulas:
1. Change the base formula: ${\log _b}M = \dfrac{{{{\log }_n}M}}{{{{\log }_n}b}}$
2. Log of a Product: ${\log _n}ab = {\log _n}a + {\log _n}b$
3. Log of a power: ${\log _a}{x^n} = n{\log _a}x$
4. Quadratic roots is given by the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
5. The square of the difference of the numbers is given by the algebraic identity :
${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$

Complete Step by Step Solution:
We are given a logarithmic function ${\log _x}36 + {\log _{18}}3x = 3$.
Now, we will change the base of the logarithmic function by using the change the base formula ${\log _b}M = \dfrac{{{{\log }_n}M}}{{{{\log }_n}b}}$, we get
$\Rightarrow \dfrac{{{{\log }_{18}}36}}{{{{\log }_{18}}x}} + {\log _{18}}3x = 3$
By using the log of a product rule ${\log _n}ab = {\log _n}a + {\log _n}b$, we get
$\Rightarrow \dfrac{{{{\log }_{18}}36}}{{{{\log }_{18}}x}} + {\log _{18}}3 + {\log _{18}}x = 3$
Rewriting in terms of exponents, we get
$\Rightarrow \dfrac{{{{\log }_{18}}{{\left( 6 \right)}^2}}}{{{{\log }_{18}}x}} + {\log _{18}}3 + {\log _{18}}x = 3$
Using the log of a power rule ${\log _a}{x^n} = n{\log _a}x$, we get
$\Rightarrow \dfrac{{2{{\log }_{18}}\left( 6 \right)}}{{{{\log }_{18}}x}} + {\log _{18}}3 + {\log _{18}}x = 3$
Now, by substituting ${\log _{18}}x = m$, we get
$\Rightarrow \dfrac{{2{{\log }_{18}}\left( 6 \right)}}{m} + {\log _{18}}3 + m = 3$
Multiplying $m$ on both the sides, we get
$\Rightarrow 2{\log _{18}}\left( 6 \right) + {\log _{18}}3m + {m^2} = 3m$
Now, by rewriting the equation, we get
$\Rightarrow 2{\log _{18}}\left( 6 \right) + {\log _{18}}3m + {m^2} - 3m = 0$
$\Rightarrow {m^2} + {\log _{18}}3 \cdot m - 3m + \left( {2{{\log }_{18}}\left( 6 \right)} \right) = 0$
Taking the terms common, we get
$\Rightarrow {m^2} + \left( {{{\log }_{18}}3 - 3} \right)m + \left( {2{{\log }_{18}}\left( 6 \right)} \right) = 0$
Now, by using the quadratic roots formula for the quadratic equation $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, we get
$m = \dfrac{{ - \left( {{{\log }_{18}}3 - 3} \right) \pm \sqrt {{{\left( {{{\log }_{18}}3 - 3} \right)}^2} - 4\left( 1 \right)\left( {2{{\log }_{18}}\left( 6 \right)} \right)} }}{2}$
Simplifying the equation, we get
$\Rightarrow m = \dfrac{{ - \left( {{{\log }_{18}}3 - 3} \right) \pm \sqrt {{{\left( {{{\log }_{18}}3 - 3} \right)}^2} - \left( {8{{\log }_{18}}\left( 6 \right)} \right)} }}{2}$
The square of the difference of the numbers is given by the algebraic identity ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$.
$\Rightarrow m = \dfrac{{ - \left( {{{\log }_{18}}3 - 3} \right) \pm \sqrt {{{\log }_{18}}^2\left( 3 \right) + 9 - 6{{\log }_{18}}\left( 3 \right) - \left( {8{{\log }_{18}}6} \right)} }}{2}$
Now, by taking the positive sign, we get
$m = \dfrac{{ - \left( {{{\log }_{18}}3 - 3} \right) + \sqrt {{{\log }_{18}}^2\left( 3 \right) + 9 - 6{{\log }_{18}}\left( 3 \right) - \left( {8{{\log }_{18}}6} \right)} }}{2}$
By simplifying the equation using calculator, we get
$\Rightarrow m = \dfrac{4}{2}$
$\Rightarrow m = 2$
Now, by taking the negative sign, we get
$m = \dfrac{{ - \left( {{{\log }_{18}}3 - 3} \right) - \sqrt {{{\log }_{18}}^2\left( 3 \right) + 9 - 6{{\log }_{18}}\left( 3 \right) - \left( {8{{\log }_{18}}6} \right)} }}{2}$
By simplifying using calculator, we get
$\Rightarrow m = 0.6199062340$
Now, by substituting $m = {\log _{18}}x$ in $m = 2$, we get
${\log _{18}}x = 2$
Now, by taking the base, we get
$\Rightarrow x = {18^2}$
Applying the exponent on the terms, we get
$\Rightarrow x = 324$
Now, by substituting $m = {\log _{18}}x$ in $m = 0.6199062340$, we get
${\log _{18}}x = 0.6199062340$
Now, by taking the base, we get
$\Rightarrow x = {18^{0.6199062340}}$
Applying the exponent on the terms, we get
$\Rightarrow x = 6$

Therefore, the solutions of $x$ for ${\log _x}36 + {\log _{18}}3x = 3$ are $6$ and $324$.

Note:
A quadratic equation is an equation that has the highest degree of 2 and has two solutions. We should be careful that the quadratic equation should be arranged in the right form. We should also notice that we have both the positive and negative signs in the formula, so the solutions for the equations would be according to the signs. We know that a logarithmic equation is an equation that involves the logarithm of an expression with a variable on either of the sides. A logarithmic is defined as the power to which number which must be raised to get some values. Logarithmic functions and exponential functions are inverses to each other.