How do you condense $ln y + ln t$ $lny+lnt$ ?

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Answer 1 · 2017-08-22T21:03:51

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

$ln y + ln t = ln (y t)$ $ln y + ln t = ln (yt)$

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

$ln a + ln b = ln (a b)$ $ln a + ln b = ln (ab)$

This follows from the corresponding property of exponents:

$e^{p + q} = e^{p} \cdot e^{q}$ $e^(p+q) = e^p * e^q$

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

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

$p = ln a$ $p = ln a" "$ and $q = ln b$ $" "q = ln b$ .

Then:

$e^{p} = a$ $e^p = a" "$ and $e^{q} = b$ $" "e^q = b$

So:

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

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

$ln y + ln t = ln (y t)$ $ln y + ln t = ln (yt)$

How do you condense lny+lntlny+lnt?