Question

y=107(4.5)^x type the model in terms of base e

Answer 1

@aum

Answer 2

@satellite73

Answer 3

@zepdrix

Answer 4

Recall that since the log and exponential function are inverses of one another:\[\Large\rm \color{orangered}{x}=e^{\ln \color{orangered}{x}}\]

And so when we have an exponential function:\[\Large\rm \color{orangered}{a^x}\]We can write it as,\[\Large\rm e^{\ln(\color{orangered}{a^x})}\]So let's apply that to our problem and see what it looks like :)

Answer 5

so what is my answer?

Answer 6

So we apply the log and exponential composition to the problem:
\[\Large\rm \color{orangered}{107(4.5)^x}=e^{\ln(\color{orangered}{107(4.5)^x})}\]And we need to apply some log rules from here so it looks a little better :)

Answer 7

ok....

Answer 8

Recall:\[\Large\rm \ln(a\cdot b)=\ln(a)+\ln(b)\]

Answer 9

So then:\[\Large\rm \ln(107(4.5)^x)=\ln(107)+\ln(4.5^x)\]

Answer 10

so is that my answer?

Answer 11

No -_- why so impatient?

Answer 12

because i really need to get this done please hurry

Answer 13

now what?

Answer 14

\[\large
4.5 = e^{\ln(4.5)} = e^{1.5041} \\ \large
\text{ } \\ \large
y=107(4.5)^x = 107(e^{1.5041})^x = 107e^{1.5041x}
\]