If 60^a=3 and 60^b=5, then what does 12^(a+b/1-b)=?

Answer 1

If 60^a=3, and 60^b=5 then, by the definition of log,
a = log[base 60] (3)
b = log[base 60] (5)

=> a + b = log[base 60] (3) + log[base 60] (5)
=> a + b = log[base 60] (3·5)
=> a + b = log[base 60] (15)

=> 1 - b = log[base 60] (60) - log[base 60] (5)
=> 1 - b = log[base 60] (60/5)
=> 1 - b = log[base 60] (12)

Now let's get this logs with base 60 onto base 12 (use the change of base formula):
log[base 60] (15) = log[base 12] (15) / log[base 12] (60)
log[base 60] (12) = log[base 12] (12) / log[base 12] (60)

so a+b/1-b =
[log[base 12] (15) / log[base 12] (60)] / [log[base 12] (12) / log[base 12] (60)] =
log[base 12] (15) / log[base 12] (12) =
log[base 12] (15) / 1 =
log[base 12] (15)

and
12^(a+b/1-b)=
12^(log[base 12] (15))=
15

Hope this helps.
________
When solving the problem, we used the following:
~ the definition of the logarithm:
x^y = a <-> a = log[base x](y)
~ the change of base formula (here from x to z):
log[base x](y) = log[base z](y) / log[base z](x)
~ the property of the sum of logs:
log[base x](y) + log[base x](w) = log[base x](y·w)
~ the property of the difference of logs:
log[base x](y) - log[base x](w) = log[base x](y/w)
~ the property of logs:
x^(log[base x](y)) = y

Answer 2

60^a=3
a = ln3/ln60

60^b=5
b = ln5/ln60

(a+b)/(1-b)
= ln15/ln12

12^[(a+b)/(1-b)]
= 12^(ln15/ln12)
= 12^log(base12)15
= 15
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It is not necessary to use base(60). Always try to get a answer in a simpler way.

Answer 3

60^a=3 and 60^b=5
a= ln3/ln60 and b = ln5/ln60
a+b = (ln3 +ln5)ln60 =ln15/ln60
1-b = 1-ln5/ln60 = (ln60-ln5)/ln 60 = ln 12/ln60
(a+b)/(1-b) = ln15/ln60 * ln60/ln12 = ln15/ln12
12 ^(ln15/ln12) = 15

Answer 4

alog60 = log3

a = log3/log60

blog60 = log5

b = log5/log60

a + b = (log3/log60 + log5/log60) = (log3 + log5)/log60

1 - b = 1 - log5/log60 = (log60 - log5)/log60

Divide

(a + b )/(1 - b) = (log3 + log5)/(log60 - log5)

(a + b )/(1 - b) = log15/log12

(a + b )/(1 - b) = log(base_12)15

12^log(base_12)15 = 15

Answer 5

first find a and b.

60^a = 3

a * log 60 = log 3

a = log 3/log 60

60^b = 5

b*log 60 = log 5

b = log 5/ log 60

determine a and b and plug in to:

12^((a+b)/(1-b))