log^a3 + log^a8 - log^a2 as a single logarithms

solve 2^(x+1) = 3^x to 2 d.p.

any help will be appreciated

billy Jan 9, 2015

#1**+10 **

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

CPhill Jan 9, 2015

#1**+10 **

Best Answer

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

CPhill Jan 9, 2015