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The Natural Base e

The Natural Base e 1.11 - Solution

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When bases are not the same, we can solve an exponential equation by taking logarithms on each side of the equation. In each of our equations, the base is the number e,e, so we will take the natural logarithm on each side. m=nln(m)=ln(n)\begin{gathered} m=n \quad \Leftrightarrow \quad \ln (m) = \ln (n) \end{gathered} Note that in order to take their logarithms, both mm and nn must be positive numbers. With this in mind, let's begin by solving the equation given in A.
9e3x=279e^{3x}=27
e3x=3e^{3x}=3
ln(LHS)=ln(RHS)\ln(\text{LHS})=\ln(\text{RHS})
ln(e3x)=ln(3)\ln \left(e^{3x}\right)= \ln (3)
ln(ea)=a \ln\left(e^a\right) = a
3x=ln(3)3x= \ln (3)
x=ln(3)3x= \dfrac{\ln (3)}{3}
x=0.3662040962x=0.3662040962\ldots
x0.366x\approx 0.366
Proceeding in the same way, we can solve the rest of the equations.
Equation Isolate Exponential Expression Take Logarithms on Each Side xx
   A. 9e3x=27\ \ \ \textbf{A. }9e^{3x}=27 e3x=3e^{3x}=3 ln(e3x)=ln(3)\ln \left(e^{3x}\right)= \ln \left(3\right) 0.366\approx 0.366
   B. 9ex=27\ \ \ \textbf{B. }9e^{x}=27 ex=3e^{x}=3 ln(ex)=ln(3)\ln \left(e^{x}\right)= \ln \left(3\right) 1.099\approx 1.099
   C. 9e3x4=27\ \ \ \textbf{C. }9e^{3x-4}=27 e3x4=3e^{3x-4}=3 ln(e3x4)=ln(3)\ln \left(e^{3x-4}\right)= \ln \left(3\right) 1.700\approx 1.700
   D. 9e3x+2=27\ \ \ \textbf{D. }9e^{3x}+2=27 e3x=259e^{3x}=\dfrac{25}{9} ln(ex)=ln(259)\ln \left(e^{x}\right)= \ln \left(\dfrac{25}{9}\right) 1.022\approx 1.022

Now we can match each equation with its solution. I: 9e3x=27iii: x0.366II: 9ex=27i: x1.099III: 9e3x4=27iv: x1.700IV: 9ex+2=27ii: x1.022\begin{aligned} \textbf{I: }& 9e^{3x}=27 \quad \quad &\textbf{iii: } x\approx 0.366 \\ \textbf{II: }& 9e^{x}=27 \quad \quad &\textbf{i: } x\approx 1.099 \\ \textbf{III: }& 9e^{3x-4}=27 \quad \quad &\textbf{iv: } x\approx 1.700 \\ \textbf{IV: }& 9e^{x}+2=27 \quad \quad &\textbf{ii: } x\approx 1.022 \end{aligned}