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kingkush85
01.10.2019 •
Mathematics
If y=4(1.6)^x, what is the approximate value of x when y=12
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Ответ:
Y = 4(1.6)^x
When y=12 solve for x
12 = 4(1.6)^x
First,we need to flip it - we need to change them sides
4(1.6)^x = 12
Divide both sides by 4
4(1.6)^x/4 = 12/4
1.6^x = 3
Now we need to solve the exponent, which is x
1.6^x = 3
To solve for x, we need to take log of both sides
log(1.6)^x = log(3)
x*(log*1.6) = log (3)
x = log(3)/log(1.6)
x = 2.34
I hope this help!
Ответ:
Or if you can't repeat, the greatest would be 98,542,100 and the smallest would be 10,024,589.