Resolve Overflow Issue: Log(K) = 127.702 | K = 5.04316 * 10^127

[r_swayze]

"If your calculator has overflow issues with this problem, you can use this identity to solve for K:

if log(K) = x + y then K = 10^x * 10^y

Problem:

log(K) = 127.702

K = 5.04316 * 10¹²⁷ "
Doesn't K = 10¹²⁷ ?

Where are they getting the 5.04316 from? How are they using that specified identity?

 

[Mark44]

log(K) = 127 + .702, so K = 10^{127 + .702}= ?

Now use the properties of exponents.

 

[HallsofIvy]

"If your calculator has overflow issues with this problem, you can use this identity to solve for K:

if log(K) = x + y then K = 10^x * 10^y

Problem:

log(K) = 127.702

K = 5.04316 * 10¹²⁷ "



Doesn't K = 10¹²⁷ ?

Where are they getting the 5.04316 from? How are they using that specified identity?

If log(K)= 127.702 then K is NOT 10¹²⁷ because K is NOT 127. You just said "log(K)= x+ y then K= 10^{x10y. 127.702= 172+ .702 so K= 1012710.702}. And 10^.702= 5.035006. (but not 5.04315- that would be closer to .703.)