Answer

Verified

374.4k+ views

**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.

Recently Updated Pages

Differentiate between Shortterm and Longterm adapt class 1 biology CBSE

How do you find slope point slope slope intercept standard class 12 maths CBSE

How do you find B1 We know that B2B+2I3 class 12 maths CBSE

How do you integrate int dfracxsqrt x2 + 9 dx class 12 maths CBSE

How do you integrate int left dfracx2 1x + 1 right class 12 maths CBSE

How do you find the critical points of yx2sin x on class 12 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Define limiting molar conductivity Why does the conductivity class 12 chemistry CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Name 10 Living and Non living things class 9 biology CBSE

The Buddhist universities of Nalanda and Vikramshila class 7 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE