# Logs and more

1. Aug 28, 2007

1. The problem statement, all variables and given/known data
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.
2. Relevant 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: $$x=\log_{a}b$$

3. 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, lets 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?

$$x=-1/2+\frac{\log_{5}12}{\log_{5}25}$$
does that make $$x=\frac{1}{2}+\frac{12}{25}$$? (wait disregard this, it doesnt matter since.. well, it doesnt ask for x)

**Genneth im trying your way but I sort of dont 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.

Last edited: Aug 28, 2007
2. Aug 28, 2007

### genneth

Your application of #6 is backwards...

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

3. Aug 28, 2007

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

4. Aug 28, 2007

### rocomath

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

5. Aug 28, 2007

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 dont think I can do that. Wait, would it be -1/2 + 1/2log base 5 of 12?

Last edited: Aug 28, 2007
6. Aug 28, 2007

### learningphysics

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

7. Aug 28, 2007

### rocomath

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

8. Aug 28, 2007

Hmm, I had gotten to that step, so then im 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?

Last edited: Aug 28, 2007
9. Aug 28, 2007

### rocomath

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

10. Aug 28, 2007

hmm

I know I can call it extraneous as logs cant be negatives, so we 'delete' or 'exclude' or 'eliminate' that one?

Last edited: Aug 28, 2007
11. Aug 28, 2007

### rocomath

yesss

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

12. Aug 28, 2007

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

13. Aug 28, 2007

### rocomath

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.

14. Aug 28, 2007

Yup, Im right.

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, Im wrong! hahaha)

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

Last edited: Aug 28, 2007
15. Aug 28, 2007

### rocomath

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

16. Aug 28, 2007

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

Last edited: Aug 28, 2007
17. Aug 28, 2007

### rocomath

you're right so far

now use the power rule and factor out a common term

18. Aug 28, 2007

### 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

19. Aug 28, 2007

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 dont 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

Last edited: Aug 28, 2007
20. Aug 28, 2007

### 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

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