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Solve for b
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Solve for b (complex solution)
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e^{-7b-3}+3=5
Use the rules of exponents and logarithms to solve the equation.
e^{-7b-3}=2
Subtract 3 from both sides of the equation.
\log(e^{-7b-3})=\log(2)
Take the logarithm of both sides of the equation.
\left(-7b-3\right)\log(e)=\log(2)
The logarithm of a number raised to a power is the power times the logarithm of the number.
-7b-3=\frac{\log(2)}{\log(e)}
Divide both sides by \log(e).
-7b-3=\log_{e}\left(2\right)
By the change-of-base formula \frac{\log(a)}{\log(b)}=\log_{b}\left(a\right).
-7b=\ln(2)-\left(-3\right)
Add 3 to both sides of the equation.
b=\frac{\ln(2)+3}{-7}
Divide both sides by -7.