Write as a single log (log is in the description)

Question

log_d(m) / log_d(2)

(basically a fraction)

Answer 1

Hello Byanca,

Assuming the question indeed asks to simplify log_d(m) / log_d(2):

You want to use the change-of-base formula for logs, but I'll use the definition of log_a(x) to break down the question.

log_a(x) = log(x)/log(a) --- #1

Now using it for your question:

log_d(m) = log(m)/log(d)

log_d(2) = log(2)/log(d)

Therefore: log_d(m) / log_d(2) = [log(m)/log(d)] / [log(2)/log(d)]

Division by a fraction means you multiply by the reciprocal of the bottom fraction so:

log_d(m) / log_d(2) = [log(m)/log(d)] * [log(d)/log(2)]

You can cross out log (d) which leaves you with:

log_d(m) / log_d(2) = [log(m)/log(2)]

Going back to equation #1, log(m)/log(2) can be expressed as log₂(m)

Cheers.

Answer 2

log₂m

One way to show that this is correct is to raise 2 to the power of both.

2^log₂m = m

2^(log_dm/log_d2) =

2^(1/log_d2)^(log_dm) =

2^(log₂d)^(log_dm) =

d^(log_dm) =

m

Answer 3

log₂m = log_dm/log_d2 where d = any base

log₂m = lnm/ln2

so if you have a calculator with just base 10 or e, you can convert a log base 2 into either as a fraction

log₂8 = 3 2³ = 8

log8/log2 = .903/.301 = 3

ln8/ln2 = ,208/.693 = 3