Question

How do you solve \[{{2}^{x+3}}={{5}^{3x-1}}\] ?

Answer 1

Hint: These types of problems can be solved by using logarithm concepts. First of all, let us consider the given equation as equation (1). Now we should apply log on both sides. Now let us consider this as equation (2). We know that \[\log {{a}^{b}}=b\log a\]. Now we should apply this concept to equation (2). Now let us consider this as equation (3). Now we can find the value of x by solving it.

Complete step by step answer:
From the given question, we were given to solve \[{{2}^{x+3}}={{5}^{3x-1}}\].
Let us consider
\[{{2}^{x+3}}={{5}^{3x-1}}....(1)\]
Now we will apply log on both sides in equation (1).
By applying log on both sides, we get
\[\Rightarrow \log {{2}^{x+3}}=\log {{5}^{3x-1}}\]
Let us consider
\[\Rightarrow \log {{2}^{x+3}}=\log {{5}^{3x-1}}.....(2)\]
We know that \[\log {{a}^{b}}=b\log a\].
Now we will apply this concept in equation (2).
By applying the above concept, we get
\[\Rightarrow (x+3)\log 2=(3x-1)\log 5\]
Let us consider
\[\Rightarrow (x+3)\log 2=(3x-1)\log 5.....(3)\]
Now from equation (3), we should find the value of x.
To find the value of x, we should have all the variables in L.H.S and all the constants in R.H.S.
From equation (3), we get
\[\begin{align}
  & \Rightarrow x\log 2+3\log 2=3x\log 5-\log 5 \\
 & \Rightarrow x\log 2-3x\log 5=-\log 5-3\log 2 \\
 & \Rightarrow x(\log 2-3\log 5)=-(\log 5+3\log 2) \\
\end{align}\]
Let us consider
\[\Rightarrow x(\log 2-3\log 5)=-(\log 5+3\log 2)......(4)\]
Now by using cross multiplication, we get
\[\begin{align}
  & \Rightarrow x=\dfrac{-(\log 5+3\log 2)}{\log 2-3\log 5} \\
 & \Rightarrow x=\dfrac{\log 5+3\log 2}{3\log 5-\log 2} \\
\end{align}\]
Let us consider
\[\Rightarrow x=\dfrac{\log 5+3\log 2}{3\log 5-\log 2}.....(5)\]
So, from equation (5), we can obtain the value of x. So, it is clear that by solving the equation \[{{2}^{x+3}}={{5}^{3x-1}}\], we get \[x=\dfrac{\log 5+3\log 2}{3\log 5-\log 2}\].

Note:
Some students may have a misconception that \[\log {{a}^{b}}=a\log b\]. But we know that \[\log {{a}^{b}}=b\log a\]. If this misconception is followed, then the final answer may get interrupted. So, these misconceptions should be avoided. Also, students should avoid calculation mistakes while solving the problem.