Natural Logarithm

IloveManUtd

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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.
 
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.
 
Hello, IloveManUtd!

I suspect the problem looks like this:

ex=2e12x+35\displaystyle e^x \:=\: 2e^{\frac{1}{2}x}+ 35


We have:   ex2e12x35  =  0\displaystyle \text{We have: }\;e^x -2e^{\frac{1}{2}x} - 35 \;=\;0

. . which equals:   (e12x)22(e12x)35  =  0\displaystyle \text{which equals: }\;\left(e^{\frac{1}{2}x}\right)^2 - 2\left(e^{\frac{1}{2}x}\right) - 35 \;=\;0


Let u=e12x\displaystyle \text{Let }\,u \,=\,e^{\frac{1}{2}x}

Substitute:   u22u35  =  0\displaystyle \text{Substitute: }\;u^2 - 2u - 35 \;=\;0

. . Factor:   (u+5)(u7)  =  0\displaystyle \text{Factor: }\;(u+5)(u-7) \;=\;0

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


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

. . e12x  =  512x=ln(-5) No real solution.\displaystyle e^{\frac{1}{2}x} \;=\;-5 \quad\Rightarrow\quad \tfrac{1}{2}x \:=\:\ln(\text{-}5) \quad\text{ No real solution.}

. . e12x=712x=ln(7)x=2ln(7)\displaystyle e^{\frac{1}{2}x} \:=\:7 \quad\Rightarrow\quad \tfrac{1}{2}x \:=\:\ln(7) \quad\Rightarrow\quad \boxed{x \:=\:2\ln(7)}
 
Solve ex = 2ex/2+35, soroban gave us the solution x = 2ln(7).\displaystyle Solve \ e^x \ = \ 2e^{x/2}+35, \ soroban \ gave \ us \ the \ solution \ x \ = \ 2ln(7).

Hence, let us do a check, x = 2ln(7) = ln(72) = ln(49).\displaystyle Hence, \ let \ us \ do \ a \ check, \ x \ = \ 2ln(7) \ = \ ln(7^2) \ = \ ln(49).

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

Hence, 49 = 2eln49+35.\displaystyle Hence, \ 49 \ = \ 2e^{ln\sqrt{49}}+35.

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

49 = 27+35\displaystyle 49 \ = \ 2*7+35

49 = 14+35\displaystyle 49 \ = \ 14+35

49 = 49, checks.\displaystyle 49 \ = \ 49, \ checks.
 
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