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

[matadorqk]

Homework Statement

Solve 2(5^{x+1})=1+\frac{3}{5^{x}} giving the answer in the form of a+\log_{5}b where a, b, are a set of integers.

Homework Equations

#1:(a^{x})(a^{y})=a^{x+y}

#2:a^{x} / a^{y}=a^{x-y}

#3:(a^{x})^{y}=a^{xy}

#4:a^{0}=1

#5:a^{1}=a

#6:\log_{a}xy=\log_{a}x+\log_{a}y

#7:\log_{a}(\frac{x}{y})=\log_{a}x-\log_{a}y

#8:\log_{a}x^{y}=y\log_{a}x

#9:\log_{a}1=0

#10:\log_{a}a=1

#11:a^{x}=b can be put as: x=\log_{a}b

The Attempt at a Solution

Here it goes:
2(5^{x+1})=1+\frac{3}{5^{x}}
So
5^{x+1}=1/2+\frac{3}{(5^{x})(2)}

Following #6(backwards), you get
x+1=\log_{5}(\frac{1}{2})(\frac{3}{(5^{x})(2)}

x+1=\log_{5}(\frac{3}{(5^{x})(4)}

so x= -1 + \log_{5}(\frac{3}{(5^{x})(4)}

so wait, let's use #7
\log_{5}3 - \log_{5}5^{x}(4)

\log_{5}3 - \log_{5}5^{x} + \log_{5}4

so \log_{5}3 - x\log_{5}5 + \log_{5}4 = \log_{5}3 - x + \log_{5}4 so it equals: 2x=-1 + \log_{5}12

How can I divide the 2?

How about this..
x=-1/2+\frac{\log_{5}12}{\log_{5}25}
does that make x=\frac{1}{2}+\frac{12}{25}? (wait disregard this, it doesn't matter since.. well, it doesn't ask for x)

**Genneth I am trying your way but I sort of don't get it

**Couple things: I put all the formulas I was given just in case, I may have skipped a few but I don't think any other is relevant. Also, I will do more than one problem in this thread, as I am.. quite stuck.. I'll do my best to minimize the number of problems, if I can solve them hhe.

 

[genneth]

Your application of #6 is backwards...

Hint: 5^(x+1) == 5*5^x. Try finding 5^x first.

 

[matadorqk]

Your application of #6 is backwards...

Hint: 5^(x+1) == 5*5^x. Try finding 5^x first.

I have no idea how that helps...:S?

 

[rocomath]

how would you get rid of the denominator on the right side?

 

[matadorqk]

how would you get rid of the denominator on the right side?

Well, it equals log_{5}25 is the same as 2 (I made it log base 5 of 25 on purpose), but the only way I can get the equation to be in the right format is if it equals '2x', and I don't think I can do that. Wait, would it be -1/2 + 1/2log base 5 of 12?

 

[learningphysics]

I have no idea how that helps...:S?

follow genneth's suggestion... then replace 5^x with a another variable say y... solve for y.

 

[rocomath]

multiply both sides by 5^x and notice that you have a product which has "the same base"

 

[matadorqk]

follow genneth's suggestion... then replace 5^x with a another variable say y... solve for y.

Hmm, I had gotten to that step, so then I am supposed to factor
10y^{2} - y - 3 = 0?

(I was never really good at factoring, so some help would be appreciated, ill try solving it anyways)

I think I factoreed correctly:
(5y-3)(2y+1)

y= 3/5
y=-1/2

Now I solve for x (y=5^x) and I use the #11 formula

So, 5^x=3/5 and 5^x=1/2
Right?

My answer is x=-1+\log_{5}3
That's for 3/5th's... for 1/2 I got x=-log_{5}2 ~ Its not right format than the one they ask, so I only give the first answer?, or both answers?

 

[rocomath]

since one of your solution equals -1/2, what can you do with that solution?

 

[matadorqk]

hmm

since one of your solution equals -1/2, what can you do with that solution?

I know I can call it extraneous as logs can't be negatives, so we 'delete' or 'exclude' or 'eliminate' that one?

 

[rocomath]

I know I can't call it extraneous, as it is not, but logs can't be negatives, so we 'delete' or 'exclude' or 'eliminate' that one?

yesss

since your calculator is log base 10, you will need to use the log-base-change formula.

 

[matadorqk]

yesss

since your calculator is log base 10, you will need to use the log-base-change formula.

Psst, read the problem, I think I am done since it says
Give the answer in a log base 5 b

 

[rocomath]

Psst, read the problem, I think I am done since it says
Give the answer in a log base 5 b

sorry :D but come on! aren't you curious to know if your answer is correct? i already know whether it is or not ;)

i'll show you the Proof right now if you want to know.

 

[matadorqk]

Yup, I am right.

sorry :D but come on! aren't you curious to know if your answer is correct? i already know whether it is or not ;)

i'll show you the Proof right now if you want to know.

Haha I guess that's part of the fun in this hehe,

So I solved for x its -.317393805 (lol)
So, (2)(5^1-.317393805)=1+3/5^-.317393805
Eventually you get 6=6 :) (Or 6.0000000005=5.999999996, I am wrong! hahaha)

**im posting another problem in a couple minutes, I am stuck, but ill try and figure it out myself first :P

 

[rocomath]

:-] i would've been stuck on that problem if genneth never posted, any more problems? I'm bored

 

[matadorqk]

:-] i would've been stuck on that problem if genneth never posted, any more problems? I'm bored

This one:
Find the exact value of x satisfying the equation
(3^{x})(4^{2x+1})=6^{x+2}

Give your answer in the form \frac{\ln a}{\ln b} where a, b are a set of integers.

Formulas I am using:
Same ones as shown above in first problem.

My solution:
Ok hows this:

(3^{x})(4^{2x+1})=6^{x+2}
So
(3^{x})(4^{2x})(4)=(36)(6^{x})
So
(3^{x})(4^{2x})=(9)(6^{x})

Now, if I multiply everything my ln would my new step look like this:
\ln(3^{x}) + \ln (4^{2x})=\ln(9) + ln (6^{x})

This is where I think I went wrong, is it right?

**Ok I am pretty sure I am right, so I continue solving, hmm..Now is right about when I may take hints perhaps?

 

[rocomath]

you're right so far

now use the power rule and factor out a common term

 

[rocomath]

the power rule states that \log_{a}b^{x} can be re-written as x\log_{a}b

so now just factor out x and solve

 

[matadorqk]

the power rule states that \log_{a}b^{x} can be re-written as x\log_{a}b

so now just factor out x and solve

umm so:
x\ln(3) + 2x\ln(4)=2ln(3)+xln(6)...
lemme just stare at that to see what I get

How am I supposed to factor x out in the right if I don't have an x in 2ln3..
I get:
x(\ln3 + 2\ln4 - \ln6)=2\ln3
So you solve inside the parenthesis
you eventually get \ln8^{x}=\ln9
Im getting somewhere.. let me think some more hehe

 

[rocomath]

on a side note, for future problems if you're solving for x

just bring the x term down, don't bother bringing the other term

x\ln{3}+x\ln{16}=\ln{9}+x\ln{6}

 

[matadorqk]

on a side note, for future problems if you're solving for x

just bring the x term down, don't bother bringing the other term

x\ln{3}+x\ln{16}=\ln{9}+x\ln{6}

Oh, right, yeah I guess its a bit simpler :p

umm.. 8^{x}=9
so x=log_{8}9
change base to e
x=\frac{\ln9}{\ln8}

YEA!

 

[rocomath]

you eventually get \ln8^{x}=\ln9
Im getting somewhere.. let me think some more hehe

you're correct, but you're solving for x so you don't need to put it back in standard form

when you solve for x such as a problem like 5x=4 you just divide, right?

it's the same for logs.

 

[matadorqk]

you're correct, but you're solving for x so you don't need to put it back in standard form

when you solve for x such as a problem like 5x=4 you just divide, right?

it's the same for logs.

:) I guess so, but now I can just go ln9/ln8, which equals 1.057

 

[rocomath]

:) I guess so, but now I can just go ln9/ln8, which equals 1.057

when verifying your answer, you want don't want to round off

this is how i check my answer on my TI with big numbers (sure i could type it all out again but screw that)

(answer) press STO-> then ALPHA then choose a letter, then just press ALPHA (letter) whenever u plug in

idk if u already do that, but just a tip, it becomes really useful in Chemistry

 

[matadorqk]

when verifying your answer, you want don't want to round off

this is how i check my answer on my TI with big numbers (sure i could type it all out again but screw that)

(answer) press STO-> then ALPHA then choose a letter, then just press ALPHA (letter) whenever u plug in

idk if u already do that, but just a tip, it becomes really useful in Chemistry

..I just started using graphic calcs.. i usually stick to my faithful casio fx-350ms
but my graphic one is fx-9750G plus.. where the hell is the STO button? lol

 

[rocomath]

i googled for an image of ur casio, lol

i found one, it's probably the arrow under the tan key

 

[matadorqk]

i googled for an image of ur casio, lol

i found one, it's probably the arrow under the tan key

Wow! Now that's awesome. I guess you learn something every day. :D

Ok let's check what problems I am missing.

I don't understand this problem:

The function f is defined for x > 2 by f(x)=\lnx+\ln(x-2)-\ln(x^{2}-4)..
(a) Express f(x) in the form of \ln(\frac{x}{x+a})
I did this, getting \ln\frac{x}{x+2} (Ill explain if you want)
(b)Find an expression for f^{-1}(x).
What the heck do they mean by that?

 

[rocomath]

so we know the log rule \log_{a}(\frac{x}{y})=\log_{a}x-\log_{a}y

so your problem is f(x)=\lnx+\ln(x-2)-\ln(x^{2}-4), now re-write it as \log_{a}(\frac{x}{y})

then you'll notice that the denominator is simply a difference of squares and you'll be able to simplify

 

[matadorqk]

so we know the log rule \log_{a}(\frac{x}{y})=\log_{a}x-\log_{a}y

so your problem is f(x)=\lnx+\ln(x-2)-\ln(x^{2}-4), now re-write it as \log_{a}(\frac{x}{y})

then you'll notice that the denominator is simply a difference of squares and you'll be able to simplify

Yeah I solved that.. I am asking what f^-1 of x is lol..

 

[rocomath]

Yeah I solved that.. I am asking what f^-1 of x is lol..

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

lol, to be honest with you ... idk how to answer that, i guess just replace the x's with y's

 

[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