Tricky logarithmic equation at International Baccalaureate high school

[physmatics]

Hi all!

From time to time, I work as a math tutor. Normally this doesn't cause me any trouble, but a couple of days ago a student from an IB school came to me with this problem:

2*5(x+1) = 1 + 3/(5x). Solve for x and write the solution on the form a + log5b.

I've been trying to solve this problem for a couple of hours, using basic logarithmic identities, but what bothers me is that I always end up with something on the form log5(a+bx) or log5(a+b*x).
I've also tried taking the logarithm on both sides, after rearranging to 2*5(x+1) - 3/(5x) = 1. This gives 0 at the right-hand side of the equation - could this be of any help?

I'm very thankful for replies!

[madah12]

(Never mind I didnt read it properly)

[eumyang]

by taking log base 5 of both sides
log(2)+(x+1) =log(3)-x
That doesn't look right.
\log_5 \left( 1 + \frac{3}{5^x} \right) = log_5 3 - x
????

No, instead, I would start by multiplying both sides by 5x, after moving all the terms to one side:
5^x \left(2\cdot 5^{x + 1} - 1 - \frac{3}{5^x} \right) = 5^x (0)
What happens next?

[physmatics]

No, instead, I would start by multiplying both sides by 5x, after moving all the terms to one side:

5^x \left(2\cdot 5^{x + 1} - 1 - \frac{3}{5^x} \right) = 5^x (0)

What happens next?

After some rearrangements, I get:
x+log_5 (2*5^{x+1}-1) = log_5 3

This is what I mean by "always ending up with something on the form log5(a+bx) or log5(a+b*x)".

Now, I could use a subtraction identity that will give me:
x+log_5 (2*5^{x+1}) + log_5 ((1-1/(2*5^{x+1}))) = log53
even though I just don't get rid of the subtraction in the logarithm. Latex stopped working for me, it should be log5(1-1/(2*5(x+1))).

This is exactly where I am stuck!

[Mentallic]

If you multiply through by 5x then you get

2.5^{x+1}.5^x=5^x+3

simplifying:

10.5^{2x}-5^x-3=0

and this is simply a quadratic in 5x, which as you can obviously tell will lead to 5^x=k and then taking the log of both sides will give you the solution for x, which you should then simplify again with log rules to get the desired form.

[eumyang]

After some rearrangements, I get:
x+log_5 (2*5^{x+1}-1) = log_5 3
I think the problem is that you're taking the logarithm of both sides "too soon." What Mentallic said was what I was trying to get you to do.

5^x \left(2\cdot 5^{x + 1} - 1 - \frac{3}{5^x} \right) = 5^x (0)

After some manipulations you should get
10 \cdot 5^{2x} - 5^x - 3 = 0
(I don't think Mentallic meant to put in decimal points.)

If you let w = 5x, then you get
10 w^{2} - w - 3 = 0
, a simple quadratic (which is factorable). After solving for w, plug back 5x in for w, and then take the logarithm of both sides.

[Mentallic]

(I don't think Mentallic meant to put in decimal points.)

Oh pssh what's the difference between my dot and the dot you used? :biggrin:

[physmatics]

Thank you all!
I get
5^{x} = 3/5
so
x = log_5 3/5
which I should have realized earlier.
Now I can be less ashamed as a tutor...

[eumyang]

Thank you all!
I get
5^{x} = 3/5
so
x = log_5 3/5
which I should have realized earlier.
Now I can be less ashamed as a tutor...
Come to think of it, are you really finished? You said in your original post:
2*5(x+1) = 1 + 3/(5x). Solve for x and write the solution on the form a + log5b.
(Emphasis mine.) If b is meant to be a positive integer, then
x = \log_5 \frac{3}{5}
is not in the form of a + log5b. But you can certainly rewrite your answer into the correct form, and I'll leave it for you to do that. :wink:

[Mark44]

Oh pssh what's the difference between my dot and the dot you used? :biggrin:

Since you're working in LaTeX anyway, you can use \cdot to get that dot so that it looks like multiplication rather than a decimal point.

[Mentallic]

Since you're working in LaTeX anyway, you can use \cdot to get that dot so that it looks like multiplication rather than a decimal point.

Ahh yeah I tend to do that kind of stuff. Like only recently did I realize that you can write \sin rather than just sin in latex and it makes it more compact and neater. Subtle differences that mean the same thing :tongue:

[HallsofIvy]

Oh pssh what's the difference between my dot and the dot you used? :biggrin:
Clearly eumyang is working at a higher level than you are! :biggrin:

[Mentallic]

Clearly eumyang is working at a higher level than you are! :biggrin:

haha :rofl:

[hunt_mat]

Write u=5^{x} to arrive at the equation:

10u=1+\frac{3}{u}\Rightarrow 10u^{2}-u-3=0

Solve this and you will get your answer, there will be some manipulation involved to get the answer 1+\log_{5}k but it will be worth it.

[hunt_mat]

This is SUCH a nice question that I am going to set my students this very question.