Question

How do you solve $5\ln x=35$ ?

Answer 1

These types of problems can be solved by considering the equation as equation (1). Then we can use the exponential formulas we can make simplify the equation. After that we have to exponentiate the equation and then by applying some basic exponential formulas we can solve the problem.

Complete step by step solution:
For the given equation, we are given to solve $5\ln x=35$ .
So let us consider the above equation as equation (1).
$5\ln x=35........\left( 1 \right)$
As we know $a\log b=\log {{b}^{a}}$ . Let us consider this formula as (f1).
Let us consider
$a\log b=\log {{b}^{a}}.........\left( f1 \right)$
By applying the formula (f1) to the equation (1) we get-
$\ln {{x}^{5}}=35$
Let us consider the above equation as equation (2).
$\ln {{x}^{5}}=35.........\left( 2 \right)$
By exponentiating both sides of equation (2), we get-
${{e}^{\ln \left( x\left( 5 \right) \right)}}={{e}^{35}}$
Let us consider the above equation as equation (3).
${{e}^{\ln \left( x\left( 5 \right) \right)}}={{e}^{35}}..............\left( 3 \right)$
As we know ${{e}^{\ln x}}=x$ . Let us consider the formula as (f2), we get-
${{e}^{\ln x}}=x...............\left( f2 \right)$
By applying the formula (f2) to equation (3), we get-
${{x}^{5}}={{e}^{35}}$
Let us consider the above equation as equation (4), we get-
${{x}^{5}}={{e}^{35}}.............\left( 4 \right)$
Now we have to take 5 from exponent, then we will get the value of x .
So by raising 1/5 th power on both sides we get-
${{\left( {{x}^{5}} \right)}^{\dfrac{1}{5}}}={{\left( {{e}^{35}} \right)}^{\dfrac{1}{5}}}$
Let us consider the above equation as equation (5).
${{\left( {{x}^{5}} \right)}^{\dfrac{1}{5}}}={{\left( {{e}^{35}} \right)}^{\dfrac{1}{5}}}.............\left( 5 \right)$
By simplifying the equation (5), we get-
$x={{e}^{^{7}}}$
Let us consider the above equation as equation (6). We get-
$x={{e}^{^{7}}}............\left( 6 \right)$
Therefore, the solution for the given problem is $x={{e}^{^{7}}}$.

Note: We can solve this problem using another method also. In that method we have to divide with 5 both sides and then by using the basic exponential formulas we can solve the problem. Students should remember that ${{e}^{\ln x}}=x$ but ${{e}^{{{\log }_{b}}a}}\ne a$ because $\ln x={{\log }_{e}}x$ . In the second case we can see that base is not equal to e that’s why the formula will not work.