Why Does Wolfram Alpha Show a Different Exponential Integration Result?

[izen]

Homework Statement

exponential integration formula $∫a^{u} du = \frac {a^{u}}{ln(a)} +c$

∫ $\left(\frac{2}{3}\right)^{x} dx$

Homework Equations

$\frac{}{}$

The Attempt at a Solution

∫ $\left(\frac{2}{3}\right)^{x} dx$

= $\frac{\left(2/3 \right)^{x}}{ln(2)/ln(3)}$ <<< this is my answer ?= - $\frac{\left(2/3 \right)^{x}}{ln(3)/ln(2)}$ << this answer from wolframalpha why ln(3)/ln(2) not ln(2)/ln(3) and where the minus comes from

please advisethank you

 

[DeIdeal]

It's just logarithm rules:

$\log{a^{b}}=b\log{a}$

Btw, you can compare if two expressions are equivalent in Wolfram with " -(2/3)^x/(ln(3/2))==(2/3)^x/(ln(2/3)) ". As you can see, the output is "True".

EDIT: I'm assuming you meant ln(3/2) and ln(2/3) instead of ln(3)/ln(2) and ln(2)/ln(3), as the former is 1) correct and 2) what Wolfram actually gives as an answer.

 

[izen]

thank you

 

[Saitama]

None of the answers you have mentioned is correct. You have done the same mistake again as you did in your previous threads.

$$\ln \frac{a}{b}≠\frac{\ln a}{\ln b}$$

 

[SammyS]

Homework Statement

exponential integration formula $∫a^{u} du = \frac {a^{u}}{ln(a)} +c$

∫ $\left(\frac{2}{3}\right)^{x} dx$

Homework Equations

Both of the following answers are incorrect.

The Attempt at a Solution

∫ $\left(\frac{2}{3}\right)^{x} dx$

= $\frac{\left(2/3 \right)^{x}}{ln(2)/ln(3)}$ <<< this is my answer ?

= - $\frac{\left(2/3 \right)^{x}}{ln(3)/ln(2)}$ << this answer from wolframalpha why ln(3)/ln(2) not ln(2)/ln(3) and where the minus comes from

please advise

thank you

Actually, WolframAlpha gives:

$\displaystyle \int \left(\frac{2}{3}\right)^{x} dx=-\frac{\left(2/3 \right)^{x}}{\ln(3/2)}+\text{constant}\ .$

As for the minus sign: $\displaystyle \ \ \ln\left(\frac{2}{3}\right)=\ln\left(\frac{3}{2} \right)^{-1}\!\!=(-1)\ln\left(\frac{3}{2}\right)\ .$

I see Pranav-Arora beat me to it!   (Way to go, P-A !)

Added in Edit:

By the way:

$\displaystyle \ln\left(\frac{2}{3}\right)=\ln(2)-\ln(3)$

On the other hand: $\displaystyle \ \ \frac{\ln(2)}{\ln(3)}=\log_{\,3}(2) \ .$

 

[izen]

None of the answers you have mentioned is correct. You have done the same mistake again as you did in your previous threads.

$$\ln \frac{a}{b}≠\frac{\ln a}{\ln b}$$

ohhh I have to be more careful about this

Thank you Pranav-Arora

Thank you Sammy for the clarification