2^a=5^b=10^c ...can someone pls help me solve this?

Question

the answer is a=(bc)/(b-c) but i cant seem to get it.can someone please show me how to do it. thanks in advance. :)

Answer 1

a=b*log2(5)=c*log2(10)
b*log2(5)=c*[log2(2)+log2(5)]
b*log2(5)=c*[1+log2(5)]
(b-c)*log2(5)=c
log2(5)=c/(b-c)
a=b*log2(5)
a=b*c/(b-c) end.

Answer 2

First, select a number to be "b".
Solve for c:

if 10^c = 5^b, then log10(10^c) = log10(5^b)
and c = b*log10(5)
(b-c) = b*(1-log10(5))

if 10^c = 2^a then log10(10^c) = log10(2^a)
and c = a*log10(2)

2^a = 2^[(bc)/(b-c)]
= 2^(c/(1-log10(5))
= 2^(c/log10(2)) (because 5*2 = 10)
= 10^c

Or, you can solve for B in terms of c and prove that 2^a = 5^b

Answer 3

2^a=5^b=10^c
Take log, a*log2=b*log5=c*log10=k (say)
Therefore, a= k/log2; b= k/log5; c= k/log10 or
log2=k/a; log5=k/b; log 10= k/c

We know from log rules, log2 +log5 = log10.
Substitute, and you get the answer

Answer 4

2^a=5^b=10^c
==> 2^a=5^b=(2^c)(5^c)
==> 2^a=(2^c)(5^c) and 5^b=(2^c)(5^c)
==> 2^(a-c)=5^c and 5^(b-c)=2^c
==> [2^(a-c)]^{c/(a-c)}=[5^c]^{c/(a-c)} and 5^(b-c)=2^c
==> 2^c =5^{c^2/(a-c)} and 5^(b-c)=2^c
hence we can equate both equation ==>5^{c^2/(a-c)}=5^(b-c)
c^2/(a-c)=b-c
==> a=c^2/(b-c)+c
= (c^2+bc-c^2)/(b-c)
= bc/(b-c) #