Exponential Equations: Solving 2(5^{x+1})=1+\frac{3}{5^{x}}

[rocomath]

lol, i actually solved it ... but idk if it's correct

y=\ln\frac{1}{x+2}

x=\ln\frac{1}{y+2}

x=-\ln{(y+2)} simplifying, ln1 = 0; dividing by -1

f^{-1}(x)=\exp^{-x}-2

 

[matadorqk]

umm

you had x/(x+2), it should be 1/(x+2) :p

1/x+2? You sure about that?

\ln x+\ln(x-2)-\ln(x^{2}-4)

\ln (x^{2}-2x)-\ln(x^{2}-4)

\ln \frac {x^{2}-2x}{x^{2}-4}

\ln \frac {(x)(x-2)}{(x+2)(x-2)}

Cancl out the "x-2"...

\ln \frac {x}{x+2}
Where does your 1 come from?

 

[rocomath]

ohhh that said lnx + ln(x-2); oops, you got it though ... argh! now the inverse is wrong (wrong if my method was correct)

 

[matadorqk]

ohhh that said lnx + ln(x-2); oops, you got it though ... argh! now the inverse is wrong (wrong if i actually did it correctly)

Well, I'll ask about the teacher how to make the inverse, I am going to move on to this:

Find \sum{ln(2^{r})} <-- (where i=1 goes on the bottom of the sum, and 50 on the top.), giving the answer in the form of a \ln (2) where a has to be a set of rational numbers.

First off, let's solve for a_{1}

\ln(2^{1})=a_{1}

Wait, its not given wether its geometric or arithmetic... umm, am I supposed to obtain that?

 

[rocomath]

Well, I'll ask about the teacher how to make the inverse, I am going to move on to this:

Find \sum{ln(2^{r})} <-- (where i=1 goes on the bottom of the sum, and 50 on the top.), giving the answer in the form of aln2 where a has to be a set of rational numbers.

First off, let's solve for a_{1}

\ln(2^{1})=a_{1}

Wait, its not given wether its geometric or arithmetic... umm, am I supposed to obtain that?

lol, i haven't done this type of problem in nearly 5 months ... i would definitely butcher this problem!

 

[matadorqk]

lol, i haven't done this type of problem in nearly 5 months ... i would definitely butcher this problem!

Lol..im missing this sum problem, another sum problem, the inverse problem, and one that hopefully you have done recently haha.

Let y=log_{3} z where z is a function of x. The diagram shows the straight line L, which represents the graph of y against x.
(a) Using the graph or otherwise, estimate the value of x when z=9
(b) The line L passes through point (1,\log_{3}\frac{5}{9}. Its gradient is 2. Find an expression for z in terms of x.

? Yeah, I have no idea. The graph you are provided goes from about (0.7, -2) to about 5.3, 6.7.. I think you won't need the graph hopefully, but any idea on how to even remotely solve this?

 

[HallsofIvy]

Well, I'll ask about the teacher how to make the inverse, I am going to move on to this:

Find \sum{ln(2^{r})} <-- (where i=1 goes on the bottom of the sum, and 50 on the top.), giving the answer in the form of a \ln (2) where a has to be a set of rational numbers.

First off, let's solve for a_{1}

\ln(2^{1})=a_{1}

Wait, its not given wether its geometric or arithmetic... umm, am I supposed to obtain that?

Why would you have to be GIVEN whether a series is geometric or arithmetic (most are neither)? Can't you check whether the definitions are satisfied? One obvious simplification here is to use the fact that
ln(2^r)= r ln(2). You series is
\sum_{i=1}^{50}r ln(2)= ln(2)\sum_{i=1}^{50} r[/itex]<br /> That last, the sum of consectutive integers, is well known.

 

[matadorqk]

Why would you have to be GIVEN whether a series is geometric or arithmetic (most are neither)? Can't you check whether the definitions are satisfied? One obvious simplification here is to use the fact that
ln(2^r)= r ln(2). You series is
\sum_{i=1}^{50}r ln(2)= ln(2)\sum_{i=1}^{50} r[/itex]<br /> That last, the sum of consectutive integers, is well known.

<br /> <br /> Right.. thanks

 

[rocomath]

remember arithmetic increases by a constant difference while geometric does not