Question

Solving the equation C = Ba^

Answer 1

@johnweldon1993 Do you know how to do this?? I don't understand.

Answer 2

Sure...so we have \[\huge C = Ba^{\frac{t}{d}} - k\] right?

Answer 3

Yes.

Answer 4

We add 'k' to both sides...because we want that (soon to be logarithm) to be alone...

So we now have

\[\huge C + k = Ba^{\frac{t}{d}}\]

Answer 5

Now we also need to divide both sides by B ...again...to isolate the soon to be logarithm...

so

\[\huge \frac{C + k}{B} = a^{\frac{t}{d}}\]

Now it is time for the logarithm...

Answer 6

\[\large a^x = x \times \log a\] so \[\huge a^{\frac{t}{d}} = \frac{t}{d}\times \log_a a\]

Answer 7

Now that is just 1 side...you need to take the log of both sides...we have

\[\huge \log_a (\frac{C + k}{B}) = \frac{t}{d} \log_a a\]

Answer 8

Now ask yourself...what is \(\large \log_a a\)? that is what power do you have to raise 'a' to to get 'a'?

Well...it's 1 ...so that just cancels out...

\[\huge \log_a (\frac{C + k}{B}) = \frac{t}{d} \]

Answer 9

Then divided by d and we have the answer, right? Omg, lol. Sorry. That was so much work I put you through. We are reviewing logs and exponentials in class and I do not for the life of me remember how to do it. I am in in Calculus right now and I haven't seen anything about logs since two years ago in Algebra 2. Thanks so much!

Answer 10

And finally...to solve for 't'...we simply divide both sides by 'd'

\[\huge t = \frac{\log_a (\frac{C + k}{B})}{d}\]

Answer 11

Haha no problem! Never worry about "putting us through work" That's what this site is about...as long as you understood it, Everything is fine!

Answer 12

Thanks again. :)