Question

How do you solve \[{{9}^{x}}={{3}^{x+4}}\]?

Answer 1

Hint: We are asked to solve the equation \[{{9}^{x}}={{3}^{x+4}}\]. To solve this, we will use the following properties, property of logarithm \[\log {{a}^{m}}=m\log a\], here \[a > 0\]. Another property is the property of exponential terms \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}\]. Also, we know that if \[c\] is square of \[d\] then, we can write \[c={{d}^{2}}\].

Complete step by step answer:
The given equation is \[{{9}^{x}}={{3}^{x+4}}\].
We know that 9 is square of 3, so we can write 9 as \[{{3}^{2}}\], using \[{{3}^{2}}\] at place of 9 in the above equation, the equation becomes \[{{\left( {{3}^{2}} \right)}^{x}}={{3}^{x+4}}\].
Using the property of exponents which states that, \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}\], we can write the above equation as \[{{3}^{2x}}={{3}^{x+4}}\].
We take the logarithm of both sides of the equation \[{{3}^{2x}}={{3}^{x+4}}\], we get
\[\Rightarrow \log \left( {{3}^{2x}} \right)=\log \left( {{3}^{x+4}} \right)\]
Using the property of logarithm which states that, \[\log {{a}^{m}}=m\log a\] we get,
\[\Rightarrow 2x\log 3=(x+4)\log 3\]
We divide both sides by \[\log 3\] we get,
\[\Rightarrow \dfrac{2x\log 3}{\log 3}=\dfrac{(x+4)\log 3}{\log 3}\]
\[\Rightarrow 2x=x+4\]
Adding \[x\] to both sides we get,
\[\begin{align}
  & \Rightarrow 2x-x=x+4-x \\
 & \therefore x=4 \\
\end{align}\]

Hence, the solution of the equation \[{{3}^{2x}}={{3}^{x+4}}\] is \[x=4\].

Note: We solved the above equation using the logarithmic method. We can also solve it by different method like the following:
The given equation is \[{{9}^{x}}={{3}^{x+4}}\].
Dividing both sides by \[{{3}^{x+4}}\], we get
\[\Rightarrow \dfrac{{{9}^{x}}}{{{3}^{x+4}}}=1\] , as we know that \[9={{3}^{2}}\], using this and the property of exponents \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}\], we can write the above equation as we get
\[\Rightarrow \dfrac{{{\left( {{3}^{2}} \right)}^{x}}}{{{3}^{x+4}}}=\dfrac{{{3}^{2x}}}{{{3}^{x+4}}}=1\]
Using the property \[\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}\] the above equation can be written as \[\Rightarrow \dfrac{{{3}^{2x}}}{{{3}^{x+4}}}={{3}^{2x-(x+4)}}={{3}^{x-4}}=1\], we know that if \[{{a}^{b}}=1\] then either \[a=1\] or \[b=0\]. Here \[a=3\], so the only possibility is for \[b=0\].
\[\Rightarrow x-4=0\]
Adding 4 to both sides, we get
\[\begin{align}
  & \Rightarrow x-4+4=4 \\
 & \therefore x=4 \\
\end{align}\]
Hence the solution for \[{{9}^{x}}={{3}^{x+4}}\] is \[x=4\].