Question

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

Answer 1

To solve this question we have to find the value of the $x$. First, we have given an equation in the exponential form so first, we change this equation in normal form by taking log both sides and using the property of logarithm and then expand that part and take variable one side and the constant part another side the put the values of basic log values and after calculating find the value of $x$.

Complete step-by-step answer:
We have given,
${3^{2 - x}} = {5^{2x + 1}}$
To find,
The value of $x$.
${3^{2 - x}} = {5^{2x + 1}}$ ……(i)
Now taking natural log both side.
$\ln \left( {{3^{2 - x}}} \right) = \ln \left( {{5^{2x + 1}}} \right)$
Now we have to use the property of logarithm $\log {x^y} = y\log x$
Using the property of logarithm
$\left( {2 - x} \right)\ln 3 = \left( {2x + 1} \right)\ln 5$
Now expanding the term by multiplying the terms.
$2\ln 3 - x\ln 3 = 2x\ln 5 + \ln 5$
Now taking the constant part one side and variable part to another side.
$2x\ln 5 + x\ln 3 = 2\ln 3 - \ln 5$
Now taking the $x$ part common
$x\left( {2\ln 5 + \ln 3} \right) = 2\ln 3 - \ln 5$
Taking the multiplied term in divide in denominator
$x = \dfrac{{2\ln 3 - \ln 5}}{{2\ln 5 + \ln 3}}$
After this, we must know the basic log values-
$x = \dfrac{{2 \times 1.098 - 1.609}}{{2 \times 1.609 + 1.098}}$
On further calculations
$x = \dfrac{{0.587}}{{4.316}}$
On dividing the terms-
$x = 0.1361$

Final answer:
The value of the x satisfying the equation ${3^{2 - x}} = {5^{2x + 1}}$ is;
$\Rightarrow x = 0.1361$

Note: Although this is a very easy question. Students must have knowledge of logarithmic identities and the rule of logarithmic functions. Students commit mistakes in using the property of logarithm and the calculations also. To find the exact value student must know all the values of logarithm with the basic base of log that is 10 and e. For higher, we calculate by using some of the properties like multiplication of log functions, the addition of log functions.