simplify log?

Question

i want to simplify 2^(log base5 7) - 7^(log base5 2)

how to start ?

Answer 1

Start by looking at base-a log b; set base-a log b = x. This means a^x = b, so ln both sides to get a ln x = ln b or x = ln(b)/ln(a).

With this, we can easily convert base-5 log 7 into ln(7)/ln(5) and base-5 log 2 into ln(2)/ln(5).

Now the problem looks like:

2^(ln 7/ln 5) - 7^(ln 2/ln 5)

Apply ln to this to get

(ln 7/ln 5) * ln 2 - (ln 2/ln 5) * ln 7

which rearranges to

(ln 7 * ln 2)/ln 5 - (ln 7 * ln 2)/ln 5 = 0.

You can check this with a calculator if you desire. 2^(base-5 log 7) = 2.311872692 and 7^(base-5 log 2) = 2.311872692.

Answer 2

Use the facts that b^log (base b) x = x, and
log(base b) N = log (base a) log (base b) a.

Use the 2nd formula to change bases from 5 to 2 and from 5 to 7. Then apply the first formula.

Then simplify.

Answer 3

2^(log base5 7) - 7^(log base5 2) =
2^(ln7/ln5) - 7^(ln2/ln5)
if x = 2^(ln7/ln5)
then
lnx = (ln7)(ln2)/ln5
x = e^[(ln7)(ln2)/ln5] (1)
if y = 2^(ln7/ln5)
lny = (ln7/ln5)ln2
y = e^[(ln7)(ln2)/ln5) (2)
so
2^(ln7/ln5) - 7^(ln2/ln5) =
e^[(ln7)(ln2)/ln5] - e^[(ln7)(ln2)/ln5) = 0