Smita04 wrote:

The positive value of x that satisfies the equation \((1 + 2x)^5 = (1 + 3x)^4\) is between

A. 0 and 0.5

B. 0.5 and 1

C. 1 and 1.5

D. 1.5 and 2

E. 2 and 2.5

OFFICIAL SOLUTION

The two expressions in question, \((1 + 2x)^5\) and \((1 + 3x)^4\), could in theory be expanded, but you’ll wind up doing a lot of algebra, only to find an equation involving \(x^5\), \(x^4\), \(x^3\), \(x^2\), and x. Solving for the positive value of x that makes the equation true is nearly impossible.

So how on Earth can you answer the question? Notice that the question does not require you to find a precise value of x; you just need a range. So plug in good benchmarks and track the value of each side of the equation.

Start with x = 1. The “equation” becomes \(3^5 =?= 4^4\).

Compute the two sides: \(3^5 = 243\), while \(4^4 = 16^2 = 256\). So the right side is bigger.

Now, we need another benchmark. Try x = 2. The “equation” becomes \(5^5 =?= 7^4\).

Compute or estimate the two sides. \(5^4 = 25^2 = 625\), and multiplying in another 5 gives you something larger than 3,000 (3,125, to be precise). Meanwhile, \(7^4 = 49^2 < 50^2 = 2,500\), so the left side is now bigger. Somewhere between x=1 and x=2, then, the equation must be true.

The only benchmark left to try is 1.5, or \(\frac{3}{2}\). Plugging in, you get \(4^5 =?= (\frac{11}{2})^4\)

\(4^5 = 2^{10} = 1,024\) (it’s good to know your powers of 2 this high)

For the other side, first compute the denominator: 24 = 16. Now the numerator: \(11^4 = 121×11×11 = 1,331×11 = 14,641\) (also quick to do longhand)

So \((\frac{11}{2})^4 = \frac{14,641}{16} < 1,000\). The right side is bigger.

Let’s recap:

\((1 + 2x)^5 = (1 + 3x)^4\) at what value of x?

\((1 + 2x)^5 < (1 + 3x)^4\) when x = 1

\((1 + 2x)^5 > (1 + 3x)^4\) when x = 1.5

\((1 + 2x)^5 > (1 + 3x)^4\) when x = 2

So the value at which the two sides are equal must be between 1 and 1.5.

The correct answer is C.

Extra points:

As x grows past 2, the larger power (5) on the left takes over, so you can see that the left side will always be bigger.

What about when x is between 0 and 1? Well, first notice that at x=0, the two sides are again equal. If x is a tiny positive number (say, 0.001 or something), then you can ignore higher powers of x (\(x^2\), \(x^3\), etc.), and this simplifies the algebraic expansion:

For tiny positive x,

\((1 + 2x)^5\) = 1 + 5(2x) + higher powers of x ? 1 + 10x

\((1 + 3x)^4\) = 1 + 4(3x) + higher powers of x ? 1 + 12x

So right away, the right side is bigger than the left side. You can also check x = ½:

\((1 + 2x)^5 = 2^5 = 32\)

\((1 + 3x)^4 = (\frac{5}{2})^4 = 2.5^4 = 6.25^2 > 36 (= 6^2) > 32\)

Again, the right side is bigger than the left side. For every x between 0 and 1, in fact, the right side is larger than the left side. This fact isn’t particularly easy to prove, but you don’t need to do so.

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