How To Really Solve This Tricky Algebra Problem? (II) - (8^x - 2^x)/(6^x - 3^x) = 2; x = ??

When you take a look at this, it does not come across as a tricky algebra problem at all. But think again. This problem is very similar to the algebra problem that I solved about a while back, in that it appears simpler on the surface (not in terms of approach to solution).

You are provided with an equation that features a fraction on the left-hand side (LHS) and a singular constant term (2) on the right-hand side (RHS). All four terms on the LHS are exponential terms with the same unknown exponent (x). Your job is now to figure out the integer value(s) for ‘x’ that would satisfy the given equation.

How would you proceed?

Spoiler Alert

If you wish to solve this equation on your own, please go ahead and do it now. Once you are done, you may choose to come back and read the rest of this essay. Beyond this point, I will be explicitly discussing the solution(s) to this equation.

This essay is supported by Generatebg

A product with a beautiful background featuring the sponsor: Generatebg - a service that generates high-resolution backgrounds in just one click. The description says "No more costly photographers" and displays a "Get Started" button beneath the description.

A Crude Approach to Solving the Tricky Algebra Problem

Let us start with a crude, yet potentially effective approach to solving this equation — the plain old trial-and-error method. Let us start by substituting ‘x’ with 0 and see what happens:

How To Really Solve This Tricky Algebra Problem? (II) — (8^x — 2^x)/(6^x — 3^x) = 2; For x = 0: (8⁰ — 2⁰)/(6⁰-3⁰) = 2; (1–1)/(1–1) = 2; 0 = 2; therefore, this is not a valid solution.
Math illustrated by the author

On the LHS, we end up with 0/0 and on the RHS, we end up with 2. In mathematics, 0/0 is ‘undefined’. If you wish to understand the reasoning behind this in detail, check out my essay on what really happens when you divide zero by zero.

In this case, when we have a defined value on the RHS (2) and an undefined value on the LHS (0/0), we have no solution. Why don’t we try our luck by substituting ‘x’ with 1 and see what comes out?

How To Really Solve This Tricky Algebra Problem? (II) — (8^x — 2^x)/(6^x — 3^x) = 2; For x = 1: (8¹ — 2¹)/(6¹-3¹) = 2; (6)/(3) = 2; 2 = 2; therefore, this is a valid solution for the given equation.
Math illustrated by the author

Bingo! We have our first valid integer solution on just our second try! This equation is perhaps not so tricky after all. Before we go off celebrating, we need to look at the bigger picture.

Sure, we have now arrived at a valid integer solution. But how do we check for other integer solutions without laboriously brute-forcing integer-substitutions into infinity? Well, we need to turn to algebra for that.

A Refined Approach to Solving the Tricky Algebra Problem

Reducing Complexity

When I look at the equation, I see potential for simplification. Often times, when we employ algebra, we aim to simplify equations to the point where we start extracting some sort of meaning out of them. In other words, we are going to try and simplify something we don’t understand into something that we do understand.

Let me start by writing out the terms in the numerator and the denominator of the LHS expressed in relation to common terms and then factor them out.

How To Really Solve This Tricky Algebra Problem? (II) — (8^x — 2^x)/(6^x — 3^x) = 2; [2^(3x) — 2^x]/[(2*3)^x — 3^x] = 2; 2^x*(2^(2x)-1)/3^x*(2^x — 1) = 2
Math illustrated by the author

Now, the terms inside the parenthesis on the numerator of the LHS represent a difference between squared terms. In such a case, we can make use of the following algebraic identity:

a² — b² = (a+b)*(a-b)

How To Really Solve This Tricky Algebra Problem? (II) — 2^x*(2^(2x)-1)/3^x*(2^x — 1) = 2 → 2^x*(2^x+1)*(2^x-1)/[3^x*(2^x-1)]= 2 (because a²-b²=(a+b)*(a-b))→ 2^x*(2^x+1)/3^x = 2
Math illustrated by the author

If we multiply both sides by 3^x, we get the following result:

How To Really Solve This Tricky Algebra Problem? (II) — 2^x*(2^x+1) = 2*3^x; Dividing by 2 on both sides → 2^(x-1)*(2^x+1) = 3^x
Math illustrated by the author

It may not be apparent yet, but this is a form that we can work with. Let’s find out how.

Analysing the Positive Integer Line

So far, we know that ‘x = 0’ is not a valid solution and ‘x = 1’ is a valid solution. If we are to look for positive integer solutions further, we need to search in the domain of ‘x≥2’. In this domain, the following situation comes into play:

How To Really Solve This Tricky Algebra Problem? (II) — 2^(x-1)*(2^x+1) = 3^x; For x≥2: LHS is even (2^(x-1) is even; (2^x+1) is odd; even times odd is even); 3^x is odd. When LHS is even and RHS is odd, we have no solution.
Math illustrated by the author

We have an even result on the LHS and an odd result on the RHS. This scenario means that there exists no other valid solution for the given equation on the positive integer line (other than ‘x = 1’).

Having come this far, the only other direction we can go in is along the negative integer line.

Analysing the Negative Integer Line

To explore the negative integer line, let us consider ‘x = -n’, where ‘n ≥ 1’. When we plug ‘-n’ in place of ‘x’, we get the following:

How To Really Solve This Tricky Algebra Problem? (II) — 2^(x-1)*(2^x+1) = 3^x; For x=-n where n≥1: 2^(x-1)*(2^x+1) = 3^x → 2^(-n-1)*(2^-n+1) = 3^-n → (1/2^(n+1))* ((1/2^n)+1) = 1/3^n → (1/2^(n+1))*[(1+2^n)/2^n] = 1/3^n → (2^n+1)/(2^(n+1)) = 1/3^n
Math illustrated by the author

If we choose to cross multiply the terms on both sides of the equation, we arrive at the following result:

How To Really Solve This Tricky Algebra Problem? (II) — (2^n+1)/(2^(n+1)) = 1/3^n; Cross-multiplying the terms on both sides → (2^n + 1)*3^n = (2^(2n+1)) → LHS is odd ((2^n + 1) is odd; 3^n is odd; odd times odd is odd) and RHS is even (2^(2n+1) is even). When LHS is odd and RHS is even, we have no solution.
Math illustrated by the author

This time, we end up with an odd term on the LHS and an even term on the RHS. This scenario means that there exists no valid solution for this equation on the negative integer line.

Therefore, we can say with certainty that the only valid integer solution for the given equation is ‘x = 1’.

Source: IMOMath

If you’d like to get notified when interesting content gets published here, consider subscribing.

Further reading that might interest you: How To Really Solve This Tricky Algebra Problem? and How To Really Benefit From Curves Of Constant Width?

If you would like to support me as an author, consider contributing on Patreon.

Street Science

Explore humanity's most curious questions!

Sign up to receive more of our awesome content in your inbox!

Select your update frequency:

We don’t spam! Read our privacy policy for more info.