# How do you condense lny+lnt?

Aug 22, 2017

Provided $y , t > 0$, we have:

$\ln y + \ln t = \ln \left(y t\right)$

#### Explanation:

Note that if $a , b > 0$ then:

$\ln a + \ln b = \ln \left(a b\right)$

This follows from the corresponding property of exponents:

${e}^{p + q} = {e}^{p} \cdot {e}^{q}$

since ${e}^{x}$ and $\ln x$ are inverses of one another.

So given $a , b > 0$, let:

$p = \ln a \text{ }$ and $\text{ } q = \ln b$.

Then:

${e}^{p} = a \text{ }$ and $\text{ } {e}^{q} = b$

So:

$\ln a + \ln b = p + q = \ln \left({e}^{p + q}\right) = \ln \left({e}^{p} \cdot {e}^{q}\right) = \ln \left(a b\right)$

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So provided $y , t > 0$ we have:

$\ln y + \ln t = \ln \left(y t\right)$