Question 891533: Solve the exponential equation using logarithms, rounding the answer to the nearest thousandth (3 decimal places). Show your work.
2 = 7x+3
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! I gather your equation is:
2 = 7^(x+3)
it's hard to figure out what it is by the way that you wrote it, but I'm taking a guess that this is the equation you wanted to show.
Assuming that it is, your solution would be as follows:
take the log of both sides of the equation to get:
log(2) = log(7^(x+3))
since log(7^(x+3)) is equivalent to (x+3)*log(7), your equation becomes:
log(2) = (x+3) * log(7)
divide both sides of this equation by log(7) and you get:
log(2) / log(7) = x + 3
solve for x + 3 using your calculator to get:
.356207... = x + 3
subtract 3 from both sides of this equation to get:
x = -2.64379...
round this to 3 decimal places and you get:
x = -2.644
That's your solution.
replace x with the rounded value of x in the original equation and you get:
2 = 7^(x+3) becomes 2 = 7^(-2.644 + 3) which becomes:
2 = 7^(.356) which becomes:
2 = 1.9991....
That's pretty close.
You would be right on if you didn't round the solution.
using the full number for x, i get:
2 = 7^(-2.643792813 + 3) becomes:
2 = 7^(.356207187 which becomes:
2 = 2
Note that .356207187 is not the full number for x + 3.
It is the number rounded to the number of decimal places that the calculator can display.
Sometimes it's right on.
Sometimes it's not.
I didn't use the displayed number.
I used the internally stored number.
That number may no even be the exact number, but, if it's not, it's at least rounded to more decimal places than the display can handle.
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