Solving for Logb(16) with Logb(2) = 0.4307

[yoleven]

Homework Statement

If logb(2)=0.4307, find logb(16)

Homework Equations

The Attempt at a Solution

If the log has the same base can I eliminate it and solve the equation?
If b^.04307=2 then b^?=16
can I say 2/.04307=16/x
16*0.4307=2x
x=3.4456
Am I close or have I missed something obvious?

 

[tiny-tim]

If logb(2)=0.4307, find logb(16)

can I say 2/.04307=16/x

NO!

Hint: 16 = 2 x 2 x 2 x 2.

 

[yoleven]

NO!

Hint: 16 = 2 x 2 x 2 x 2.

Okay, 2^4=16
If I have b^.4307=2 By trial and error, I came up with 5 for b. 5^.4307=2 or log5(.4307)=2
Specifically, what steps do I follow to discover what the base is without resorting to a trial and error method.
Thanks

 

[Avodyne]

Start with b^.4307=2, and raise both sides to a certain power, such that you will get 16 on the right.

 

[tiny-tim]

If logb(2)=0.4307, find logb(16)

Okay, 2^4=16
If I have b^.4307=2 By trial and error, I came up with 5 for b. 5^.4307=2 or log5(.4307)=2
Specifically, what steps do I follow to discover what the base is without resorting to a trial and error method.

Hi yoleven!

You don't need to find b … the question doesn't ask you for b.

Hint: 16 = 2 x 2 x 2 x 2.

logb(pq) = logb(p) + logb(q)

 

[HallsofIvy]

Or, more simply for this problem log(a^b)= b log(a).

 

[tiny-tim]

Or, more simply for this problem log(a^b)= b log(a).

oooh … that's far too advanced!

 

[yoleven]

If logb(2)=.4307
logb(16)=1.7228
because if b^.4307=2, (b^.4307)^4=(2)^4
b^1.7228=16
Okay? Thanks again.

 

[FordPrefect]

1.7228 is correct.