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Question: How do you solve \( 7 = 5{e^{0.2x}} \) ?...

How do you solve 7=5e0.2x7 = 5{e^{0.2x}} ?

Explanation

Solution

Hint : In order to determine solution of the given equation, simply it first by dividing both sides by 55 .Now We will convert the expression into logarithmic form, and to do so use the definition of exponents that the exponent of the form by=x{b^y} = x is when converted into logarithmic form is equivalent to logbX=y{\log _b}X = y ,so compare with the given logarithm value with this form . And write it into logarithmic form and solve the equation for xx to get the required result.

Complete step-by-step answer :
We are given an exponential equation in variable xx
7=5e0.2x7 = 5{e^{0.2x}}
Simplifying the equation by dividing both sides of the equation with 55 , we get
15×7=15×5e0.2x 75=e0.2x   \dfrac{1}{5} \times 7 = \dfrac{1}{5} \times 5{e^{0.2x}} \\\ \dfrac{7}{5} = {e^{0.2x}} \;
Now we are going to convert the above exponential form into logarithmic form.
Any exponential form by=X{b^y} = X when converted into equivalent logarithmic form results in logbX=y{\log _b}X = y
So in our case we have 75=e0.2x\dfrac{7}{5} = {e^{0.2x}} ,comparing it with by=X{b^y} = X , we get values of variables as
X=75 b=e y=0.2x   X = \dfrac{7}{5} \\\ b = e \\\ y = 0.2x \;
So we have logarithmic form of our equation as
loge75=0.2x\Rightarrow {\log _e}\dfrac{7}{5} = 0.2x
Since logarithm with base as ee is simply written as ln\ln and known as” natural logarithm”.
ln75=0.2x\Rightarrow \ln \dfrac{7}{5} = 0.2x
Now dividing both sides of the equation with 0.20.2 , we have value of xx as
10.2ln75=0.2x0.2 5ln75=x x=5ln75   \Rightarrow \dfrac{1}{{0.2}}\ln \dfrac{7}{5} = \dfrac{{0.2x}}{{0.2}} \\\ \Rightarrow 5\ln \dfrac{7}{5} = x \\\ \Rightarrow x = 5\ln \dfrac{7}{5} \;
Therefore, the solution of the given equation is x=5ln75x = 5\ln \dfrac{7}{5} .
So, the correct answer is “x=5ln75x = 5\ln \dfrac{7}{5}”.

Note : 1. A logarithm is basically the reverse of a power or we can say when we calculate a logarithm of any number , we actually undo an exponentiation.
2.Any multiplication inside the logarithm can be transformed into addition of two separate logarithm values .
logb(mn)=logb(m)+logb(n){\log _b}(mn) = {\log _b}(m) + {\log _b}(n)
3. Any division inside the logarithm can be transformed into subtraction of two separate logarithm values .
logb(mn)=logb(m)logb(n){\log _b}(\dfrac{m}{n}) = {\log _b}(m) - {\log _b}(n)
4. Any exponent value on anything inside the logarithm can be transformed and moved out of the logarithm as a multiplier and vice versa.
nlogm=logmnn\log m = \log {m^n}