Solving the equation C = Ba ^ (t/D) - K for t using logarithms with base a gives you:
A. D loga (C/B +K)
B. loga (D(C+K)/B)
C. 1/D loga (C/B) +K
D. (loga ((C+K)/B)) / D
E. D loga ((C+K)/B)
I was able to solve up to (t/D) lna = (C+K)/B
actually, since the base is a, not e,
t/D = log_a (C+K)/B
t = D log_a ((C+K)/B)
so, choice E
oobleck is right just did it on APEX Calculus
True Im on apex too thats correct
Well, well, looks like you're trying to solve an equation using logarithms. Math can be a real clown fiesta sometimes, but fear not, I'm here to help you make sense of it!
Alright, so you've got (t/D) lna = (C+K)/B, and you want to find t. Let's break it down together.
First, let's isolate t by multiplying both sides of the equation by D:
(t/D) lna * D = (C+K)/B * D
The D on the left side cancels out, giving us:
t lna = D(C+K)/B
Now, we need to get t by itself. We can do this by dividing both sides of the equation by lna:
t = (D(C+K)/B) / lna
Aha! Now we've got t all alone. But hey, we can simplify this further using logarithm properties. Remember that ln(a/b) is equal to ln(a) - ln(b). So we can rewrite the equation as:
t = D(C+K)/(Blna)
And there you have it! The answer is option E. D loga ((C+K)/B). Keep on clowning around with those logarithms, my friend!
To solve the equation C = Ba^(t/D) - K for t using logarithms with base a, you've correctly arrived at the equation (t/D) lna = (C+K)/B.
To isolate t and solve for it, you need to multiply both sides of the equation by D and divide by lna:
(t/D) lna = (C+K)/B
Multiply both sides by D:
t lna = (C+K)D/B
Divide both sides by lna:
t = (C+K)D/(B lna)
So, the correct answer is option E: D loga ((C+K)/B).