Question

Solve ${9^{1 + \log x}} - {3^{1 + \log x}} - 210 = 0$, where base of log is $3$.

Answer 1

Hint: We will write this equation in terms of ${3^{1 + \log x}}$ and then we will assume ${3^{1 + \log x}} = m$. We will get a quadratic equation and then we will solve it for the values of m and hence the value of ${3^{1 + \log x}}$. We will then take logarithms both sides with base $3$ and hence solve it for the value of x.

Complete step-by-step answer:
 We are given an equation: ${9^{1 + \log x}} - {3^{1 + \log x}} - 210 = 0$ with base $3$.
We are required to solve this equation or in other words, we have to solve it for the value of x.
We can write this equation: ${9^{1 + \log x}} - {3^{1 + \log x}} - 210 = 0$ as:
$ \Rightarrow {\left( 3 \right)^{\left( 2 \right)1 + \log x}} - {3^{1 + \log x}} - 210 = 0$
Or, ${\left( {{3^{1 + \log x}}} \right)^2} - {3^{1 + \log x}} - 210 = 0$
Let ${3^{1 + \log x}} = m$. Now the equation becomes:
$ \Rightarrow {m^2} - m - 210 = 0$
We get a quadratic equation in m. Let us solve it for the value of m by the factorization method as:
$ \Rightarrow {m^2} - 15m + 14m - 210 = 0$
$ \Rightarrow m\left( {m - 15} \right) + 14\left( {m - 15} \right) = 0$
$ \Rightarrow \left( {m + 14} \right)\left( {m - 15} \right) = 0$
$ \Rightarrow m = - 14{\text{ or }}15$
Here, negative value is not acceptable since ${3^{1 + \log x}}$ is positive always.
Therefore, $m = 15$.
$ \Rightarrow {3^{1 + \log x}} = 15$
Using the base change rule ${\log _b}a = x \Leftrightarrow {b^x} = a$, we can write this equation as:
$ \Rightarrow {\log _3}15 = 1 + {\log _3}x$
$ \Rightarrow {\log _3}x = {\log _3}15 - 1$
Now, we know that ${\log _x}x = 1$. Using this formula, we can write the equation as:
$ \Rightarrow {\log _3}x = {\log _3}15 - {\log _3}3$
Using the logarithmic formula: ${\log _a}m - {\log _b}n = {\log _a}\left( {\dfrac{m}{n}} \right)$ in the above equation, we get
$ \Rightarrow {\log _3}x = {\log _3}\left( {\dfrac{{15}}{3}} \right) = {\log _3}5$
$ \Rightarrow {\log _3}x = {\log _3}5$
Taking anti – logarithm both sides, we get
$ \Rightarrow $$x = 5$

Note: In this question, we may get confused at many steps especially where we have supposed ${3^{1 + \log x}} = m$ since this way it becomes easier to solve for the solution. Also, when we have used the base change rule of the logarithms. Here, we have compared ${3^{1 + \log x}} = 15$ with ${b^x} = a$ and after that we have used the base change formula and it was already provided in the question itself that the base of the log is $3$.