Question

How do you solve \[{2^{3x}} = 3{e^x}\] ?

Answer 1

Hint: Here in this question, we have to solve the expression for x variable. To solve this we have to take log on both sides and rearrange the expression by using the properties of logarithms and further simplification by using basic arithmetic operations, we get the required value of x variable.

Complete step-by-step answer:
Consider the given expression:
 \[ \Rightarrow {2^{3x}} = 3{e^x}\] -------(1)
Here, x is the variable, we have to solve the expression for the x variable.
Take a log on both side in equation (1), we have
 \[ \Rightarrow \ln {2^{3x}} = \ln 3{e^x}\] ------(2)
Now using the, some logarithmic properties:
Product property:
If a, m and n are positive integers and \[a \ne 1\] , then;
 \[{\log _a}\left( {mn} \right){\text{ }} = {\text{ }}{\log _a}m{\text{ }} + {\text{ }}{\log _a}n\]
Power rule or property:
If a and m are positive numbers, \[a \ne 1\] and n is a real number, then;
Rearrange the LHS by power rule and RHS by product property in equation (2), we have
 \[ \Rightarrow 3x\ln 2 = \ln 3 + \ln {e^x}\] ---------(3)
As we know the logarithm function with base e become \[{\log _e}e = 1\] , \[\ln \] is the natural log with base e, then equation (3) becomes
 \[ \Rightarrow 3x\ln 2 = \ln 3 + x\]
As by using the log calculator the value of \[\ln 2 = 0.69314718056\] and \[\ln 3 = 1.09861228867\] on substituting these values in equation (4), then
 \[ \Rightarrow 3x\left( {0.69314718056} \right) = 1.09861228867 + x\]
 \[ \Rightarrow 2.07944154168x = 1.09861228867 + x\]
Subtract x on both side, then
 \[ \Rightarrow 2.07944154168x - x = 1.09861228867\]
 \[ \Rightarrow 1.07944154168x = 1.09861228867\]
Now we have to solve for x
 \[ \Rightarrow x = \dfrac{{1.09861228867}}{{1.07944154168}}\]
On simplification, we get
 \[ \Rightarrow x = 1.01775987513\]
On round off, we get
 \[1.01775987513\;\, \cong 1.02\]
This is the required answer.
So, the correct answer is “1.02”.

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.