Algebra: How to solve for b when $\ln ab^t = -0.12t + 4.67$?

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I'm trying to solve the following for b:

$\ln ab^t = -0.12t + 4.67$ where $-0.12t + 4.67$ is the equation of a straight line.

I apply basic log rules:

$\ln a + t \ln b = -0.12t + 4.67$

According to the mark scheme / answer, you can just compare the coefficients of 't':

$-0.12 = \ln b$

But I don't understand, doesn't that imply the following:

$4.67 = \ln a$

If so, how can we be sure when there are two constants, a and b?

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using the logarithm rules we get $$\ln(a)+t\ln(b)=-\frac{3}{25}t+\frac{467}{100}$$ and this is equivalent to $$\ln(b)=\frac{1}{t}\left(-\frac{3}{25}t+\frac{467}{100}-\ln(a)\right)$$ thus $$b=e^{\frac{1}{t}\left(-\frac{3}{25}t+\frac{467}{100}-\ln(a)\right)}$$