Question 71864: Can you help with the following?
A*(10^5)^B=140*10^6
A*(10^7)^B=100*10^6
Find A and B
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Can you help with the following?
A*(10^5)^B=140*10^6
A*(10^7)^B=100*10^6
Find A and B
:
Using ordinary logs would be helpful, (powers of 10 are the logs):
The 1st equation:
log(A) + 5B = log(140) + 6
log(A) + 5B = 2.146128 + 6
log(A) + 5B = 8.146128
:
The 2nd equation:
log(A) + 7B = log(100) + 6
log(A) + 7B = 2 + 6
log(A) + 7B = 8
:
Subtract the 1st equation from the 2nd equation, that eliminates log(A):
log(A) + 7B = 8
log(A) + 5B = 8.146128
----------------------------subtract
2B = - .146128
B = -.146128/2
B = -.073 (rounded off, these decimal places are getting tedious)
:
Find log(A) using; log(A) + 7B = 8, substitute -.073 for B:
log(A) + 7(-.073) = 8
log(A) + -.511 = 8
log(A) = 8 + .511
log(A) = 8.511
A = 324,339,617, an intimidating number, let's see if works in the original 1st equation.
:
A*(10^5)^B = 140*(10^6)
324,339,617 * (10^5)^-.073 = 140*(10^6)
324,339,617 * .43152 = 140(10^6)
139,959,031 ~ 140(10^6), close but I should not have rounded off so much
:
Try it in the 2nd equation:
324339617(10^7)^-.073 = 100(10^6)
324339617 * .30832 = 100(10^6)
99,999,999.89 ~ 100(10^6) close enough
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