1. ## weird log proof

got his as a homework question and not sure if it is a misprint or this is the proper equation. Either way, got no clue how to proceed.

2^x = 5^y = 10^z
using logarithms to base 10 or otherwise, show that 1/z = 1/x + 1/y

i am thinking one of those equal signs is really a plus maybe a minus???

any help greatly appreciated.

jacs

2. Originally Posted by jacs
got his as a homework question and not sure if it is a misprint or this is the proper equation. Either way, got no clue how to proceed.

2^x = 5^y = 10^z
using logarithms to base 10 or otherwise, show that 1/z = 1/x + 1/y

i am thinking one of those equal signs is really a plus maybe a minus???

any help greatly appreciated.

jacs
Nope. The problem statement is correct. Let's use log base 10:

2^x = 10^z

log(2^x) = log(10^z) = z* log(10) = z

x*log(2) = z

or 1/x = log(2)/z

Similarly: 1/y = log(5)/z

Thus
1/x + 1/y = log(2)/z + log(5)/z = [log(2) + log(5)]/z = log(2*5)/z = log(10)/z = 1/z (Check)

-Dan

3. Originally Posted by jacs
got his as a homework question and not sure if it is a misprint or this is the proper equation. Either way, got no clue how to proceed.

2^x = 5^y = 10^z
using logarithms to base 10 or otherwise, show that 1/z = 1/x + 1/y

i am thinking one of those equal signs is really a plus maybe a minus???

any help greatly appreciated.

jacs
Say that,
2^x=t
5^y=t
10^z=t
Then,
x log 2=log t
y log 5=log t
z=log t
Thus,
1/z=1/log t
1/x=log 2/log t
1/y=log 5/log t

But,
1/x+1/y=log 5/log t+log 2/log t=(log 2+log 5)/log t
Note the numerator is the identity for logarithm product,
log (2*5)/log t=log 10/log t=1/log t=1/z

This is mine 26th post!!!