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.
_________________
"Be challenged at EVERY MOMENT."“Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.”"Each stage of the journey is crucial to attaining new heights of knowledge."Rules for posting in verbal forum | Please DO NOT post short answer in your post!
Advanced Search : https://gmatclub.com/forum/advanced-search/