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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

Best Answer 

 #1
avatar+87301 
+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
avatar+87301 
+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

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