Graph of exponential function

Yuseph

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Yo guys,

Look Im trying to answer question c : find x when y = 1,6.
I know the answer is x = - 0,74. Because its in the book and because my graphing calculator says so. But i have no idea how to get there using only algebra. The author skipped that part.
As youll see in the picture I made 2 pathetic attempts using logarithm but there re two problems. Firstly in chapter logarithm there was never an exercice that showed taking out only half the exponent and switching to lhs of log. And im too new to know if its possible or not. Secondly if i proceed otherwise and take out both 0.3 and x then e is left alone and i dont think theres such a thing as e alone without exponent. Now it could be that its not about logarithm at all. I really dont know.
 

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Yo guys,

Look Im trying to answer question c : find x when y = 1,6.
I know the answer is x = - 0,74. Because its in the book and because my graphing calculator says so. But i have no idea how to get there using only algebra. The author skipped that part.
As youll see in the picture I made 2 pathetic attempts using logarithm but there re two problems. Firstly in chapter logarithm there was never an exercice that showed taking out only half the exponent and switching to lhs of log. And im too new to know if its possible or not. Secondly if i proceed otherwise and take out both 0.3 and x then e is left alone and i dont think theres such a thing as e alone without exponent. Now it could be that its not about logarithm at all. I really dont know.
y = 2 * e0.3*x

y/2 = e0.3*x

lne(0.8) = 0.3*x

-0.223143551 = 0.3 * x

x = -0.7438
 
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This is not a very efficient approach, but is mostly valid, except for one major error. The right-hand side is 2 times a power of e, but you are taking it as a power of 2e, particularly when you get to the fourth line. You have violated the order of operations.

I see that you have been shown the efficient method of solving, so I'll just describe it. The general idea is to isolate an exponential before taking a log; that's why I would first divide by 2. (This also reflects the order of operations, and makes the next step easier to do correctly.)

Then, when you have an exponential on one side, it is most efficient to use that base, in this case e, for the logarithm. That is not necessary, but appropriate. (If the base is not 10 or e, and the goal is a decimal answer, then you can just use either base 10 or base e for the sake of calculating.)
 
Thanks. Yea i thought the 2 was mulyiplying only e. Now I can see it doesnt matter if only a part of the exponent is taken. Both ways work. Also when the equation include e then ln is the quickest and cleanest way.
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I like when people try different methods to discover for themselves why other's recommendations are good (or not).

You may also notice that the differences in your answers are due to rounding errors. When I do the calculations using all the digits in the calculator, rather than typing rounded values back in, I get -0.74381183771403251922098363436612 in each case.
 
Yea thanks. I really need to know all the details or I cant sleep.
The natural logarithm exercices came right after actually. They re easy.
But im stuck with sth new right now.
Until now ive learnt that if sth is shifted on lfs you do the same on rhs. I also learnt that whether its an expression or an equation when you reverse a fraction that has a exponent the sign of the exponent change to the opposite sign. But here i have this exercice i have no idea how the author did it. Sometimes he just does his thing and take the explanation for granted.
Look at the picture. How on earth did he move from 9/4 = e^ -3x to 4/9 = e^3x ?
I solved it another way because i cant follow something i dont understand.
 

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As he says, he takes the reciprocal of both sides, which is what you describe as "reversing" a fraction (also called inverting, or, informally, "flipping"). The reciprocal is the same as dividing 1 by a number: the reciprocal of x is 1/x, which is the same as x^-1.

The reciprocal of 9/4 is 4/9. The reciprocal of a^n is a^-n, because 1/a^n is a^-n. So the reciprocal of e^(-3x) is e^(3x).

But your work is fine; the reciprocal is just a little trick to get rid of the negative.
 
Yea thanks. I really need to know all the details or I cant sleep.
The natural logarithm exercices came right after actually. They re easy.
But im stuck with sth new right now.
Until now ive learnt that if sth is shifted on lfs you do the same on rhs. I also learnt that whether its an expression or an equation when you reverse a fraction that has a exponent the sign of the exponent change to the opposite sign. But here i have this exercice i have no idea how the author did it. Sometimes he just does his thing and take the explanation for granted.
Look at the picture. How on earth did he move from 9/4 = e^ -3x to 4/9 = e^3x ?
I solved it another way because i cant follow something i dont understand.
Do you realize:

\(\displaystyle \frac{9}{4} = \frac{1}{\left(\frac{4}{9}\right)}\)

\(\displaystyle e^{-3x}\ =\ \frac{1}{e^{3x}}\)
 
Right so the reciprocal is not necessarely flipping a fraction. Good to know.
No i thought the inversed fraction was 1/e(3)(-x).
Because e^3x = 9/4 gives e^(3)(-0,27033) = 9/4 which gives 2,25 = 2,25
Now if i take the reciprocal 1/ e^3x = 4/9 it gives 1/e^(3)(-0,27033) = 4/9 which results in 2,25 = 0,444. Thats why i typed 1/(3)(-x) instead. But then i dunno ive never that case before
 
Right so the reciprocal is not necessarely flipping a fraction. Good to know.
No i thought the inversed fraction was 1/e(3)(-x).
Because e^3x = 9/4 gives e^(3)(-0,27033) = 9/4 which gives 2,25 = 2,25
Now if i take the reciprocal 1/ e^3x = 4/9 it gives 1/e^(3)(-0,27033) = 4/9 which results in 2,25 = 0,444. Thats why i typed 1/(3)(-x) instead. But then i dunno ive never that case before

I'm not sure what you are saying. The equation is e^{-3x} = 9/4 , not e^{3x} = 9/4. [ Please observe that the exponent needs grouping symbols; I used { } here. Your e^(3)(-0,27033) would mean e^3, multiplied by (-0,27033). You need e^{(3)(-0,27033)}.]

When we evaluate e^{-3(-0.27033)}, we get e^0.81099 = 2.25 as required. Your e^{(3)(-0,27033)} is e^-0.81099 = 0.4444, that is, 4/9, not 9/4.

The reciprocal of e^{-3x} can be written as 1/e^{-3x} or e^{3x}. It is not 1/e^{3x}.
 
Thanks guys. Yea i got lost in explanation. What i mean is the only reciprocal i dealt with so far is the one on lhs in the picture. Never done it with an x exponent involved. The piece of information missing for my problem was the non flipping inversion. Nice trick. I guess i have all the pieces now.
 

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