Solve for b
b=-q+\ln(28)
Solve for q
q=-b+\ln(28)
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e^{b+q}+2=30
Use the rules of exponents and logarithms to solve the equation.
e^{b+q}=28
Subtract 2 from both sides of the equation.
\log(e^{b+q})=\log(28)
Take the logarithm of both sides of the equation.
\left(b+q\right)\log(e)=\log(28)
The logarithm of a number raised to a power is the power times the logarithm of the number.
b+q=\frac{\log(28)}{\log(e)}
Divide both sides by \log(e).
b+q=\log_{e}\left(28\right)
By the change-of-base formula \frac{\log(a)}{\log(b)}=\log_{b}\left(a\right).
b=\ln(28)-q
Subtract q from both sides of the equation.
e^{q+b}+2=30
Use the rules of exponents and logarithms to solve the equation.
e^{q+b}=28
Subtract 2 from both sides of the equation.
\log(e^{q+b})=\log(28)
Take the logarithm of both sides of the equation.
\left(q+b\right)\log(e)=\log(28)
The logarithm of a number raised to a power is the power times the logarithm of the number.
q+b=\frac{\log(28)}{\log(e)}
Divide both sides by \log(e).
q+b=\log_{e}\left(28\right)
By the change-of-base formula \frac{\log(a)}{\log(b)}=\log_{b}\left(a\right).
q=\ln(28)-b
Subtract b from both sides of the equation.
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