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

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Last updated date: 15th Jun 2024
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Answer
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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.

Complete step by step answer:
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.