Question

How do you solve ${3^{2x + 1}} = {5^{x + 2}}$ ?

Answer 1

Hint : Here we have to solve the above equation. The above equation is in the form of exponential form. The exponential number is defined as the number of times we multiply the number by itself. So, we can’t solve this directly so we apply log to it and solve further.

Complete step-by-step answer :
The logarithmic function and the exponential function are both inverse of each other. The exponential number can be written in the form of a logarithmic number and likewise we can write the logarithmic number in the form of an exponential number.
Now we have equation ${3^{2x + 1}} = {5^{x + 2}}$
Apply log on the both sides we have

$$\Rightarrow \log \left( {{3^{2x + 1}}} \right) = \log \left( {{5^{x + 2}}} \right) \\\ \Rightarrow \log {3^{2x + 1}} = \log {5^{x + 2}} \;$$

By the logarithmic property we have $\log {a^m} = m\log a$ , using this property we have
$\Rightarrow (2x + 1)\log 3 = (x + 2)\log 5$
We multiply the terms we have
$\Rightarrow 2x\log 3 + \log 3 = x\log 5 + 2\log 5$
Let we group the terms which involve the variable x as one term and the remaining one term. So take $x\log 5$ to LHS and log3 to RHS we have
$\Rightarrow 2x\log 3 - x\log 5 = 2\log 5 - \log 3$
Take x as a common in the LHS of the above equation
$\Rightarrow x(2\log 3 - \log 5) = 2\log 5 - \log 3$
Now we write the equation for x so we have
$\Rightarrow x = \dfrac{{2\log 5 - \log 3}}{{2\log 3 - \log 5}}$
In the numerator and denominator the first term can be written as
$\Rightarrow x = \dfrac{{\log {5^2} - \log 3}}{{\log {3^2} - \log 5}}$
On simplification we have
$\Rightarrow x = \dfrac{{\log 25 - \log 3}}{{\log 9 - \log 5}}$
By using the Clark’s table finding the each term value and then on simplification we get
$\Rightarrow x = 3.607$ , nearly.
So, the correct answer is “x = 3.607”.

Note : The exponential number is inverse of logarithmic. But here we have not used this. We have applied the log on both terms. The logarithmic functions have several properties on addition, subtraction, multiplication, division and exponent. So we have to use logarithmic properties. We have exact values for the numerals by the Clark’s table, with the help of it we can find the exact value.