Question

help! the integral of 4^(3x) dx

Answer 1

I keep getting 4^3x/ 3ln(4) +C

Answer 2

but that's not an answer choice

Answer 3

\(\color{black}{\displaystyle y=a^x}\)
\(\color{black}{\displaystyle \ln y=\ln a^x}\)
\(\color{black}{\displaystyle \ln y=x \ln a}\)
\(\color{black}{\displaystyle y'/y= \ln a}\)
\(\color{black}{\displaystyle y'= y\ln a}\)
\(\color{black}{\displaystyle y'= a^x\ln a}\)

Since the derivative of \(a^x\) is \(a^x\ln a\), therefore,
the antiderivative of \(a^x\) is \((a^x/\ln a)\color{gray}{+C}\).

So, you may conclude,
\(\color{black}{\displaystyle \int a^x~dx=\frac{a^x}{\ln a} +C:\quad \quad \left\{\left.a\right|a>0,~a\ne e\right\}}\)

Answer 4

Well, a can also be equal to e, actually....
e^x/ln(e) = e^x/1 = e^x,

which just repeats that
integral of e^x is just e^x

Answer 5

try using properties of log @chris215

Answer 6

maybe like the power rule

Answer 7

\[\ln(x^r)=? \text{ thing power rule for log}\]

Answer 8

\(\color{black}{\displaystyle \int b^{cx}~dx=\int (b^c)^{x}~dx=\frac{(b^c)^x}{\ln (b^c)} +C=\frac{b^{cx}}{c\ln (b)} +C}\)

Answer 9

this is another derivation from my previous derivation, using a property
\(a^{bc}=(a^b)^c\)

Answer 10

I got 4^(3x)/(ln4)+C

Answer 11

ln(x^r)=r ln(x)

Answer 12

so 3 ln(4)=?

Answer 13

6ln2?

Answer 14

so it would be 4^(3x)/6ln2 +c ?

Answer 15

you could write 3 ln(4) as 6 ln(2)

Answer 16

6•ln(2) = 4•ln(3)

Yes

Answer 17

is that one of your choices ?

Answer 18

oh the other way,
3 ln(4)

Answer 19

yes, you are right.

Answer 20

yes it is thanks so much !

Answer 21

lol I didn't even notice with my dislexic-ness that what you wrote was different than what I wrote @idku

Answer 22

but after much concentration i finally seen it

Answer 23

:)