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# Find the value of x, if ${9^{1 + \log x}} - {3^{1 + \log x}} - 210 = 0$, the base is $3$.

Last updated date: 15th Jun 2024
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Hint: We have to find the value of $x$. For this, first, we need to simplify the given equation.
To solve the equation, we will apply a few formulae as follows:
${a^{m + n}} = {a^m}.{a^n}$
${a^{{{\log }_a}x}} = x$
We know that, if the multiplication of two terms is zero, the value of each term is individually zero.

It is given that, ${9^{1 + \log x}} - {3^{1 + \log x}} - 210 = 0$ the base is $3$.
We have to find the value of $x$.
Here, we have,
$\Rightarrow {9^{1 + \log x}} - {3^{1 + \log x}} - 210 = 0$
We know that,
$\Rightarrow {a^{m + n}} = {a^m}.{a^n}$
Applying the formula, we get,
$\Rightarrow {9.9^{\log x}} - {3.3^{\log x}} - 210 = 0$
Simplifying we get,
$\Rightarrow {9.3^{2\log x}} - {3.3^{\log x}} - 210 = 0$
Since, the base is $3$. We have, ${3^{\log x}} = x$
So, applying the formula we get,
$\Rightarrow 9{x^2} - 3x - 210 = 0$
Dividing each term by $3$ we get,
$\Rightarrow 3{x^2} - x - 70 = 0$
Now we will split the middle term as follows,
$\Rightarrow 3{x^2} - 15x + 14x - 70 = 0$
Simplifying we get,
$\Rightarrow 3x(x - 5) + 14(x - 5) = 0$
Simplifying again we get,
$\Rightarrow (x - 5)(3x + 14) = 0$
We know that, if the multiplication of two terms is zero, the value of each term is individually zero.
So, we have,
$\Rightarrow (x - 5) = 0$ gives $x = 5$ and
$\Rightarrow (3x + 14) = 0$ gives $x = \dfrac{{ - 14}}{3}$
We only take positive value for $x$ as the logarithm of negative numbers is not defined.

$\therefore$ The value of $x$ is $5$.

Note:
In Quadratic Factorization using Splitting of Middle Term which is $x$ term is the sum of two factors and product equal to the last term.
Logarithm, the exponent or power to which a base must be raised to yield a given number. Expressed mathematically, $x$ is the logarithm of $n$ to the base $b$ if ${b^x} = n$, in which case one writes $x = {\log _b}n$.
When rewriting an exponential equation in log form or a log equation in exponential form, it is helpful to remember that the base of the logarithm is the same as the base of the exponent.