Question

17/8 = e^(ln7/8 t)

here, don't e and ln cancel out each other?

Answer 1

Yes.

Answer 2

can u try solving for t?

Answer 3

because when i cancel them out, i got t= 2.42, which is different from the answer

Answer 4

ans given is -5.64

Answer 5

i would not call that "cancellation" but yes, \(e^{\ln(x)}=x\)

Answer 6

Is the contents of the log \(\dfrac{7}{8t}\) or \(\dfrac{7}{8}t\)?

Answer 7

the 2nd one

Answer 8

Hmm then your solution sounds correct... that's strange.. thinking :3

Answer 9

the solution apples an extra ln

Answer 10

OH I SEE!
The t is NOT inside the log, that's the issue here.

Answer 11

So you can't "cancel" out the e and log, because the t is in the way.

Answer 12

\[\huge e^{t \cdot \ln\frac{7}{8}}\]See the problem?

Answer 13

oh, so u can separate it as e^t + e^ln7/8 -> then here u can change to 7/8?

Answer 14

Yah that's another way to approach it! :)

Answer 15

and still need to apply ln to isolate t

Answer 16

hah thanks zepdirx :D