Answer
427.8k+ views
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.
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.
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