Part b)
90 = 6 e^(1.18 t )
15 = e^(1.18 t )
ln 15 = 1.18 t ln e
ln 15 = 1.18 t
t = ln 15 / 1.18
t = 2.29 months

b)

M(t) is the mice’s population, therefore if the population is 90, M(t) = 90

So, 90 = 6e^(1.18t)
e^(1.18t) = 15

since, 15 is a constant function, it is continuous on R (real numbers); and e^(1.18t) is defined on R and by definition continuous on it,
we can put the LN (natural log) in both side of the equality,

ln(e^(1.18t)) = ln(15)
hence, 1.18t = ln(15) {because, ln is the inverse function of e, in other words (e o ln)(x) = (ln o e)(x) = x }

then, 1.18t = 2.71 (approximately)

Finally, t = 2.29
Basically, in a bit more than two months there will be 90 mice

if u wanna make it more accurate (but i guess there is no real need)
1 month –> 30 days
0.29 month –> ?
?=0.29*30 = 8.7

so in 2 months and about 9 days there will be 90 mice……

(b)
M(t) = 6e ^ 1.18t
90=6e^1.18t
15=e^1.18t

Note that ln is the natural logarithm meaning log base e

ln15=ln(e)^1.18t
ln15=1.18t ln(e)
ln15=1.18t
(ln15)/1.18=t
t = (ln15)/1.18
That’s the answer but you’ll be needing the calculator to get the exact value.

M(0) = 6e ^1.18 *0 = 6e^0 = 6

(recall that any number to the power of zero is one).

b) for M(t) = 90;

6e ^ 1.18t = 90

e ^ 1.18t = 15

Take natural log on both sides;

1.18t = ln (15)

t = ln (15) / 1.18

And your problem is solved!

b)M(t) = 6e ^ 1.18t = 90
e^1.18t = 90/6
1.18t = ln(90/6)
t = (ln(90/6))/1.18