- #1
jokerzz
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I think I know how to do this but I'm completely blank at the moment. Do I differentiate 'ln' and then differentiate 2^x and 3^x in the denominator?
bigubau said:Hint
[tex] \lim_{n\rightarrow\infty} \frac{\ln\left(3^n\left(1+\frac{2^n}{3^n}\right)\right)}{n} [/tex]
Err.. Sorry, but I couldn't get your point... What are you trying to do?
And now use the logarithm's properties.
jokerzz said:ok so that worked but when I take 2^n common i get the answer "ln(2)+ln(1.5)" but wen i take 3^n common, i get "ln(3)+ln(2/3)" and these are different answers
bigubau said:You';re making a mistake somewhere.
[tex] \lim_{n\rightarrow\infty} \frac{2^n}{3^n} = 0 [/tex]
bigubau said:Hint
[tex] \lim_{n\rightarrow\infty} \frac{\ln\left(3^n\left(1+\frac{2^n}{3^n}\right)\right)}{n} [/tex]
And now use the logarithm's properties.
VietDao29 said:May I ask, what does taking this limit have anything to do with differentiating y = ln(3x + 2x)?
bigubau said:It's the same thing. 2/3 is under unity. Multiplying it an infinite nr of times gives 0.
The OP wanted to differentiate that in order to use L'Hopital's rule to do that limit. As usual, there was a far easier way to find the limit than to use L'Hopital's rule. Another case of not asking the question you really want answered!VietDao29 said:May I ask, what does taking this limit have anything to do with differentiating y = ln(3x + 2x)?
The formula for differentiating ln(2^x+3^x) is d/dx ln(2^x+3^x) = (2^x ln2 + 3^x ln3) / (2^x + 3^x).
To simplify ln(2^x+3^x), you can use the logarithm rules to rewrite it as ln(2^x) + ln(3^x). Then, using the power rule for logarithms, you can further simplify it to xln2 + xln3.
The derivative of ln(2^x+3^x) is (2^x ln2 + 3^x ln3) / (2^x + 3^x).
Yes, you can use the chain rule to differentiate ln(2^x+3^x). The first step is to rewrite it in the form ln(u) where u = 2^x+3^x. Then, the derivative is d/dx ln(u) = 1/u * du/dx = (2^x ln2 + 3^x ln3) / (2^x + 3^x).
The critical points of ln(2^x+3^x) occur when the derivative is equal to 0 or undefined. In this case, the derivative is undefined when x=0, and it is equal to 0 when x=-ln(ln3/ln2). Therefore, the critical points are x=0 and x=-ln(ln3/ln2).