Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 months ago

# 7. Find a number t such that ln(2t + 1) = −4 exact value?

precalc

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• 2 months ago

ln(2t+1)=-4

=>

2t+1=e^(-4)

=>

2t=e^(-4)-1

=>

2t=(1-e^4)/e^4

=>

t=(1-e^4)/(2e^4)

• 2 months ago

ln(2t + 1) = -4

Making each side an exponent over a base of "e" will cancel out the natural log and leave only what's inside.  Doing that we get:

2t + 1 = e⁻⁴

Now solving for t:

2t = e⁻⁴ - 1

t = (e⁻⁴ - 1) / 2

or:

t = 1/(2e⁴) - 1/2

That's your exact value.  A decimal approximation to that would be:

t = -0.4908422

Substituting this into the original equation we can test this value:

ln(2t + 1) = -4

ln[2(-0.4908422) + 1] = -4

ln(-0.9816844 + 1) = -4

ln(0.0183156) = -4

-4.000002 = -4

Since we rounded to 7SF, we should be correct to 6SF.  Checking the first 6 SF of the left side it does equal the right side.  So this is correct.

t = 1/(2e⁴) - 1/2

• fcas80
Lv 7
2 months ago

t can be solved in terms of e.  e is irrational.  The resulting expression will be exact in terms of e, but it will not be an integer.

• 2 months ago

ln(2t + 1) = -4

so, 2t + 1 = e⁻⁴

Hence, 2t = e⁻⁴ - 1

so, t = (e⁻⁴ - 1)/2

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