How To Really Solve This Tricky Algebra Problem (IX) - Whiteboard style graphics showing the following information: 2^(3x) − 2^x = 336; x = (what)??; where x is a real number.

Welcome to the ninth entry in the tricky algebra problem series. Following the last entry, we are still dealing with a relatively easy yet tricky puzzle here. You are presented with the following equation:

[2^(3x)] − (2^x) = 336

where x ∈ R

Given this setting, your challenge is to solve for ‘x’. Unlike some of the previous puzzles that I have covered in this series, this one actually presents no hidden challenges; it is as easy as it looks. Now that you know, how would you solve this problem?

Spoiler Alert

I will be explicitly discussing solutions to this problem beyond this point. So, if you wish to solve it on your own, I’d recommend that you pause reading at this point and proceed with your attempt.

Once you’ve completed your attempt, you may choose to continue reading the essay to compare your approach with mine.

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Setting Up the Tricky Algebra Problem and Simplifying the Equation

As soon as I glance at the given equation, I see that both terms on the left-hand side have ‘x’ as a common exponent. To make our lives easier, why don’t we define a parameter ‘t’ as follows:

t = 2^x

Consequently, the original equation transforms as follows:

How To Really Solve This Tricky Algebra Problem (IX) — Whiteboard style graphics showing the following information: t = 2^x; Therefore, t³ − t = 336
Parametrised equation — Math illustrated by the author

We may further simplify this equation by factoring out ‘t’ on the left-hand side as follows:

How To Really Solve This Tricky Algebra Problem (IX) — Whiteboard style graphics showing the following information: t³ − t = 336; → t*(t² − 1) = 336; → t*(t + 1)*(t − 1) = 336 [Because, (a² − b²) = (a + b)*(a − b)]
Simplified equation — Math illustrated by the author

If you look closely, you will notice that this equation directly leads us to at least one of the roots we are looking for. Can you see the pattern? If not, do not worry. Let us find out what is going on here.

Unravelling the Pattern and Solving for ‘t’

Let us rearrange the multiplicative terms on the left-hand side as follows:

(t − 1)*t*(t + 1) = 336

If we consider only integers (which are a subset of real numbers), we see immediately that the left-hand side asserts multiplication of three consecutive integers.

Furthermore, the right-hand side features just a 3-digit number. What this means is that just by trying out a few combinations of numbers below 9, we are likely to land on our first root. Why don’t we do just that:

2*3*4 = 24

3*4*5 = 60

4*5*6 = 120

5*6*7 = 210

6*7*8 = 336 → Bingo!

We have just established that 7 is a root for the parametrised expression we are solving. So, the resulting expression would look as follows:

How To Really Solve This Tricky Algebra Problem (IX) — Whiteboard style graphics showing the following information: (t − 7)*(some expression) = t³ − t − 336 = 0
Reworked equation (unsolved) — Math illustrated by the author

The next step becomes a simple process of reverse-engineering what the missing expression should be so that the rest of the equation remains valid.

We have t³ on the right-hand side. So, surely, the missing expression must have t² in it. Similarly, one after the other, we may deduce the terms of the missing expression. The final result is as follows:

How To Really Solve This Tricky Algebra Problem (IX) — Whiteboard style graphics showing the following information: (t − 7)*(some expression) = t³ − t − 336 = 0; → (t − 7)*(t² + 7t + 48) = 336
Reworked equation (solved) — Math illustrated by the author

We may find the remaining roots by solving the quadratic expression which we have now deduced. But before we do that, let us check if it has real roots to begin with. In algebra, we do this by calculating the discriminant (Δ) of the expression.

How To Really Solve This Tricky Algebra Problem (IX) — Whiteboard style graphics showing the following information: Consider: t² + 7t + 48; Δ = b² − (4*a*c) = 49 − (4*1*48) = −143 < 0
Delta calculation to see if there are real roots — Math illustrated by the author

Since the discriminant is less than zero, the expression (t² + 7t + 48) has no real roots. So, we can say for certain that (t = 7) is the only possible root.

But before we celebrate, remember that we have just solved the parametrised equation. We still need to work out the solution in terms of ‘x’.

The Solution for the Tricky Algebra Problem

We know that t = 2^x. So, to solve for ‘x’, we need to solve the following equation:

2^x = 7

Like I have done in the past, my intuitive approach with such equations is to first ‘dethrone’ the unknown exponent.

To do this, let us employ the logarithm. That is, we can solve the above equation by taking the natural logarithm of both the sides as follows:

How To Really Solve This Tricky Algebra Problem (IX) — Whiteboard style graphics showing the following information: ln(2^x) = ln(7); Applying the power rule of logarithm: x*ln(2) = ln(7); Dividing both sides by ln(2): x = ln(7)/ln(2); x = 2.80735 (approximately);
The solution to the tricky algebra problem — Math illustrated by the author

There we go. This is the solution to our tricky algebra problem in this entry. I hope you had as much fun solving this problem as I did!

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