Natural Logarithm

[IloveManUtd]

How do you solve: e^x = 2e^1/2x + 35. I know I have to subsitute a special number but what do I do with the power of 1/2? Please help. Thanks.

 

[mmm4444bot]

We need to type parentheses around exponents, when exponents are more than a single number or variable.

I think the exponent is x/2 + 35.

e^x = 2 e^(x/2 + 35)

Take the natural logarithm of both sides.

ln(e^x) = ln(2 e^[x/2 + 35])

x = ln(2) + ln(e^[x/2 + 35])

x = ln(2) + x/2 + 35

2x = 2 ln(2) + x + 70

You do the last step.

 

[soroban]

Hello, IloveManUtd!

I suspect the problem looks like this:

\(\displaystyle e^x \:=\: 2e^{\frac{1}{2}x}+ 35\)

\(\displaystyle \text{We have: }\;e^x -2e^{\frac{1}{2}x} - 35 \;=\;0\)

. . \(\displaystyle \text{which equals: }\;\left(e^{\frac{1}{2}x}\right)^2 - 2\left(e^{\frac{1}{2}x}\right) - 35 \;=\;0\)


\(\displaystyle \text{Let }\,u \,=\,e^{\frac{1}{2}x}\)

\(\displaystyle \text{Substitute: }\;u^2 - 2u - 35 \;=\;0\)

. . \(\displaystyle \text{Factor: }\;(u+5)(u-7) \;=\;0\)

. . \(\displaystyle \text{and we have: }\;u \;=\;-5,\:7\)


\(\displaystyle \text{Back-substitute:}\)

. . \(\displaystyle e^{\frac{1}{2}x} \;=\;-5 \quad\Rightarrow\quad \tfrac{1}{2}x \:=\:\ln(\text{-}5) \quad\text{ No real solution.}\)

. . \(\displaystyle e^{\frac{1}{2}x} \:=\:7 \quad\Rightarrow\quad \tfrac{1}{2}x \:=\:\ln(7) \quad\Rightarrow\quad \boxed{x \:=\:2\ln(7)}\)

 

[BigGlenntheHeavy]

\(\displaystyle Solve \ e^x \ = \ 2e^{x/2}+35, \ soroban \ gave \ us \ the \ solution \ x \ = \ 2ln(7).\)

\(\displaystyle Hence, \ let \ us \ do \ a \ check, \ x \ = \ 2ln(7) \ = \ ln(7^2) \ = \ ln(49).\)

\(\displaystyle Ergo, \ e^{ln(49)} \ = \ 2e^{ln(49)/2}+35.\)

\(\displaystyle Hence, \ 49 \ = \ 2e^{ln\sqrt{49}}+35.\)

\(\displaystyle 49 \ = \ 2e^{ln(7)}+35\)

\(\displaystyle 49 \ = \ 2*7+35\)

\(\displaystyle 49 \ = \ 14+35\)

\(\displaystyle 49 \ = \ 49, \ checks.\)