How do you solve 1-2e^(2x)=-191−2e2x=−19?

3 Answers
Dean R.
May 1, 2018

x = ln \sqrt{10} x=ln√10

Explanation:

1 - 2 e^{2x } = -19 1−2e2x=−19

-2 e^{2x} = -19 -1 = -20 −2e2x=−19−1=−20

e^{2x} = -20/(-2) = 10 e2x=−20−2=10

ln e^{2x} = ln 10 lne2x=ln10

2x = ln 10 2x=ln10

x = {ln 10}/2 = ln \sqrt{10} x=ln102=ln√10

Check:

1 - 2 e^{2x } 1−2e2x

= 1 - 2 e^{2 (ln sqrt{10} ) } =1−2e2(ln√10)

= 1 - 2 e^{ln 10} =1−2eln10

= 1 - 2(10) =1−2(10)

= -19 quad sqrt

Manasvini Nittala
May 1, 2018

the value is ~~1.151

Explanation:

given 1-2e^(2x)=-19rArr-2e^(2x)=-20rArre^(2x)=10
in general we have e^m=krArr log_ek=m
which means we have log_e10=2x and log_e10~~2.302
we have 2x=2.302rArrx~~1.151

Karthik N
May 1, 2018

x = (ln10)/2
~~1.1512925465

Explanation:

Subtract 1 on both sides.
-2e^(2x) = -20
Divide by -2.
e^(2x) = 10
Taking the logarithm of both sides, we have:
ln(e^(2x)) = ln10
Using the power rule of logarithms,
2xln(e) = ln 10

lne = 1 So, we have:

2x = ln 10
x = (ln10)/2

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