log^a3 + log^a8 - log^a2 as a single logarithms
I assume this is supposed to be
loga^3 + loga^8 - loga^2 if so, we have...
log [ (a^3 * a^8) / a^2) ] =
log ( a^11 / a^2) =
log a^9 =
9log(a)
------------------------------------------------------------------------------------------------------------------------
solve 2^(x+1) = 3^x to 2 d.p.
Take the log of both sides
log 2^(x + 1) = log 3^x and by a property of logs, we have
(x + 1) log 2 = x log3 simplify
x log 2 + log 2 = x log3 rearrange
x log 2 - x log 3 = - log 2 factor
x (log 2 - log 3) = - log 2 divide both sides by (log 2 - log 3)
x = (-log 2) /(log2 - log3) = about 1.71