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two questions about 'ln' on C3

When I get an equation of 'f(x)= ln(2+3x)', how do I derive it from the original equation of 'ln x'.

Am I right in thinking that to get from 'ln x' to 'f(x)= ln(2+3x)', I divide the x values of 'ln x' by three, and then shift the whole curve 2 units to the left ? Also, where would the asymtote be in 'f(x)= ln(2+3x)'?



my other question is, when find I was finding the inverse of 'ln(5x-2)', I got: 'e^(x+5) / 5'. But in the answers, it has exponential just to the power of 'x', whereas I thought it would be to the power of 'x+5'. Can someone explain this for me?
W.H.T
When I get an equation of 'f(x)= ln(2+3x)', how do I derive it from the original equation of 'ln x'.

Am I right in thinking that to get from 'ln x' to 'f(x)= ln(2+3x)', I divide the x values of 'ln x' by three, and then shift the whole curve 2 units to the left ? Also, where would the asymtote be in 'f(x)= ln(2+3x)'?



my other question is, when find I was finding the inverse of 'ln(5x-2)', I got: 'e^(x+5) / 5'. But in the answers, it has exponential just to the power of 'x', whereas I thought it would be to the power of 'x+5'. Can someone explain this for me?


Yes, remember, stretches and reflections first, translations last. The asymptote would be when x = -2/3 because that makes the equation ln0 which is undefined.

For the inverse of ln(5x-2)
The steps are simple so I might as well just show you:

y = ln(5x-2)
x = ln(5y-2)
e^x = 5y - 2

5y = e^x + 2
y = (e^x +2)/5
1. yup u r right, the asymtote will be at -2/3 i think

2. I got the inverse as (e^x + 2)/5 if its right than I can explain :biggrin:

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