Question

How do you evaluate $5{{e}^{3x+7}}=21$?

Answer 1

Hint: To evaluate the given expression $5{{e}^{3x+7}}=21$, divide both sides by 5 and to undo the exponential function with a base of $e$, take the logarithm of both sides with base $e$. Remember that ${{\log }_{e}}\left( x \right)$is the natural logarithm, denoted by $\ln \left( x \right)$ and apply one of the properties of natural logarithm that \[\ln \left( {{e}^{u\left( x \right)}} \right)=u\left( x \right)\]. Simplify it further using addition, subtraction and log table values.

Complete step by step solution:
Given equation in the question is:
$5{{e}^{3x+7}}=21$
To simplify it further Divide this equation with \[5\]both the sides, we get:
\[\Rightarrow {{e}^{3x+7}}=\dfrac{21}{5}\]
And to undo the exponential function with a base of $e$, take the logarithm of both sides with base $e$that is $\ln \left( x \right)$, we get:
\[\Rightarrow \ln \left( {{e}^{3x+7}} \right)=\ln \left( \dfrac{21}{5} \right)...\left( 1 \right)\]
Apply one of the properties of natural logarithm that is \[\ln \left( {{e}^{u\left( x \right)}} \right)=u\left( x \right)\] we get
\[\Rightarrow \ln \left( {{e}^{3x+7}} \right)=3x+7...\left( 2 \right)\]
Now putting the value of equation \[\left( 2 \right)\] in equation \[\left( 1 \right)\] we get:
\[\Rightarrow 3x+7=\ln \left( \dfrac{21}{5} \right)\]
Subtract \[7\] from both the sides from the above equation we get:
\[\Rightarrow 3x=\ln \left( \dfrac{21}{5} \right)-7\]
To simplify it further Divide this equation with \[3\] both the sides, we get:
\[\Rightarrow x=\dfrac{\ln \left( \dfrac{21}{5} \right)-7}{3}\]
By using a log table, we can find the value of \[\ln \left( \dfrac{21}{5} \right)\approx 1.4350\]. Now substituting this value in above equation, we get:
\[\Rightarrow x\approx \dfrac{1.4350-7}{3}\]
\[\Rightarrow x\approx \dfrac{-5.5649}{3}\]
On simplifying it further we get:
\[\Rightarrow x\approx -1.8550\]
Therefore, in the given equation $5{{e}^{3x+7}}=21$ the value of \[x\] is \[-1.8550\].

Note: It’s important to remember that ${{\log }_{e}}\left( x \right)$is the natural logarithm, denoted by $\ln \left( x \right)$ and apply one of the properties of natural logarithm that \[\ln \left( {{e}^{u\left( x \right)}} \right)=u\left( x \right)\].
 Alternative way to solve this question is as follows:
Given equation in the question is:
$5{{e}^{3x+7}}=21$
Now we take the logarithm of both sides with base $e$that is $\ln \left( x \right)$, we get:
\[\Rightarrow \ln \left( 5{{e}^{3x+7}} \right)=\ln \left( 21 \right)...\left( 1 \right)\]
Apply one of the properties of natural logarithm that is \[\ln \left( u\left( x \right)v\left( x \right) \right)=\ln \left( u\left( x \right) \right)+\ln \left( v\left( x \right) \right)\] we get:
\[\Rightarrow \ln \left( 5{{e}^{3x+7}} \right)=\ln \left( 5 \right)+\ln \left( {{e}^{3x+7}} \right)...\left( 2 \right)\]
Now putting the value of equation \[\left( 2 \right)\] in equation \[\left( 1 \right)\] we get:
\[\Rightarrow \ln \left( 5 \right)+\ln \left( {{e}^{3x+7}} \right)=\ln \left( 21 \right)\]
\[\Rightarrow \ln \left( {{e}^{3x+7}} \right)=\ln \left( 21 \right)-\ln \left( 5 \right)\]
Now apply one of the properties of natural logarithm that is \[\ln \left( u\left( x \right) \right)-\ln \left( v\left( x \right) \right)=\ln \left( \dfrac{u\left( x \right)}{v\left( x \right)} \right)\] we get:
\[\Rightarrow \ln \left( {{e}^{3x+7}} \right)=\ln \left( \dfrac{21}{5} \right)...\left( 3 \right)\]
Apply one of the properties of natural logarithm that \[\ln \left( {{e}^{u\left( x \right)}} \right)=u\left( x \right)\] we get
\[\Rightarrow \ln \left( {{e}^{3x+7}} \right)=3x+7...\left( 4 \right)\]
Now putting the value of equation \[\left( 4 \right)\] in equation \[\left( 3 \right)\] we get:
\[\Rightarrow 3x+7=\ln \left( \dfrac{21}{5} \right)\]
Subtract \[7\] from both the sides from the above equation we get:
\[\Rightarrow 3x=\ln \left( \dfrac{21}{5} \right)-7\]
To simplify it further Divide this equation with \[3\] both the sides, we get:
\[\Rightarrow x=\dfrac{\ln \left( \dfrac{21}{5} \right)-7}{3}\]
By using a log table, we can find the value of \[\ln \left( \dfrac{21}{5} \right)\approx 1.4350\]. Now substituting this value in above equation, we get:
\[\Rightarrow x\approx \dfrac{1.4350-7}{3}\]
\[\Rightarrow x\approx \dfrac{-5.5649}{3}\]
On simplifying it further we get:
\[\Rightarrow x\approx -1.8550\]