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Hint- We need to solve the given problem using basic formulae of logarithms.

Given logarithmic equation is ${\log _x}\left( {3{x^2} + 10x} \right) = 3$

$ \Rightarrow 3{x^2} + 10x = {x^3}$ $\left[ {\because {{\log }_a}b = x \Leftrightarrow b = {a^x}} \right]$

Simplifying the above equation, we get

$ \Rightarrow {x^3} - 3{x^2} - 10x = 0$

Factorization of the above equation give

$ \Rightarrow {x^3} + 2{x^2} - 5{x^2} - 10x = 0$

$ \Rightarrow ({x^2} - 5x)(x + 2) = 0$

$ \Rightarrow x(x - 5)(x + 2) = 0$

The domain of x is x>0, since x=0 and x=-2 do not lie in the domain of x.

$ \Rightarrow x = 5$ is the only solution of the given equation.

Note:

In the given problem our goal is to find x value. We have x in the base of logarithm and in the logarithmic expression. We used the formula of equivalence in logarithms $\left[ {\because {{\log }_a}b = x \Leftrightarrow b = {a^x}} \right]$ , to take x out from the base. Then we got a polynomial equation of degree 3. We solved for x value. We got three values for x out of those one is negative, one is zero and the remaining one is positive value. Logarithmic bases should be always greater than zero. So the positive value is the required x value.

Given logarithmic equation is ${\log _x}\left( {3{x^2} + 10x} \right) = 3$

$ \Rightarrow 3{x^2} + 10x = {x^3}$ $\left[ {\because {{\log }_a}b = x \Leftrightarrow b = {a^x}} \right]$

Simplifying the above equation, we get

$ \Rightarrow {x^3} - 3{x^2} - 10x = 0$

Factorization of the above equation give

$ \Rightarrow {x^3} + 2{x^2} - 5{x^2} - 10x = 0$

$ \Rightarrow ({x^2} - 5x)(x + 2) = 0$

$ \Rightarrow x(x - 5)(x + 2) = 0$

The domain of x is x>0, since x=0 and x=-2 do not lie in the domain of x.

$ \Rightarrow x = 5$ is the only solution of the given equation.

Note:

In the given problem our goal is to find x value. We have x in the base of logarithm and in the logarithmic expression. We used the formula of equivalence in logarithms $\left[ {\because {{\log }_a}b = x \Leftrightarrow b = {a^x}} \right]$ , to take x out from the base. Then we got a polynomial equation of degree 3. We solved for x value. We got three values for x out of those one is negative, one is zero and the remaining one is positive value. Logarithmic bases should be always greater than zero. So the positive value is the required x value.

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