How do you condense #lny+lnt#?

Question

1662 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-22T21:03:51

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

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

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

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

This follows from the corresponding property of exponents:

#e^(p+q) = e^p * e^q#

since #e^x# and #ln x# are inverses of one another.

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

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

Then:

#e^p = a" "# and #" "e^q = b#

So:

#ln a + ln b = p + q = ln(e^(p+q)) = ln(e^p * e^q) = ln(ab)#

#color(white)()#
So provided #y, t > 0# we have:

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