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/(5^x). 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+b^x) or log₅(a+b*x).
I've also tried taking the logarithm on both sides, after rearranging to 2*5^(x+1) - 3/(5^x) = 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 5^x, 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+b^x) 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 5^x 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 5^x, 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 = 5^x, then you get
$$10 w^{2} - w - 3 = 0$$
, a simple quadratic (which is factorable). After solving for w, plug back 5^x 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?

 

[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/(5^x). 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.

 

[Mark44]

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

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?

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

 

[Mentallic]

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

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.