Question

⑫(11 + x)³⁄₂ = 27

Explanation:

Step1: Rewrite the equation

The given equation is \((11 + x)^{\frac{3}{2}} = 27\). We can rewrite \(27\) as \(3^3\) and also rewrite the left - hand side exponent in a more familiar form. Recall that \(a^{\frac{m}{n}}=\sqrt[n]{a^m}\), so \((11 + x)^{\frac{3}{2}}=\sqrt{(11 + x)^3}\) or we can also think of it as \(((11 + x)^{\frac{1}{2}})^3\). Let's first raise both sides of the equation to the power of \(\frac{2}{3}\) to eliminate the exponent \(\frac{3}{2}\) on the left - hand side.

Using the property of exponents \((a^m)^n=a^{mn}\), if we have \(((11 + x)^{\frac{3}{2}})^{\frac{2}{3}}=(27)^{\frac{2}{3}}\)

Step2: Simplify both sides

For the left - hand side, \(((11 + x)^{\frac{3}{2}})^{\frac{2}{3}}=11 + x\) (by the rule \((a^m)^n=a^{m\times n}\), here \(m = \frac{3}{2}\) and \(n=\frac{2}{3}\), so \(\frac{3}{2}\times\frac{2}{3}=1\))

For the right - hand side, we simplify \(27^{\frac{2}{3}}\). First, \(27 = 3^3\), so \((3^3)^{\frac{2}{3}}\). Again, using the rule \((a^m)^n=a^{m\times n}\), we have \(3^{3\times\frac{2}{3}}=3^2 = 9\)

So now our equation becomes \(11+x=9\)? Wait, no, that can't be right. Wait, maybe I misread the original equation. Wait, looking back at the original hand - written equation, maybe it's \((11 + x)^{\frac{3}{2}}=27\) or maybe it's \((11 + x)^{\frac{3}{2}} = 27\), but when we raise \(27\) to the \(\frac{2}{3}\) power:

Another way to calculate \(27^{\frac{2}{3}}\):

First, find the cube root of \(27\), \(\sqrt[3]{27}=3\), then square the result: \(3^2 = 9\). And for the left - hand side, \(((11 + x)^{\frac{3}{2}})^{\frac{2}{3}}=11 + x\) (since \((a^{\frac{m}{n}})^{\frac{n}{m}}=a\) when \(a\geq0\) and \(m,n\in\mathbb{Q},n
eq0\))

Wait, but if \(11 + x=9\), then \(x=9 - 11=- 2\), but let's check:

If \(x=-2\), then \(11+x = 9\), and \(9^{\frac{3}{2}}=\sqrt{9^3}=\sqrt{729}=27\), which works. Wait, but let's do it again step by step.

Starting over:

Given \((11 + x)^{\frac{3}{2}}=27\)

Raise both sides to the \(\frac{2}{3}\) power:

\(((11 + x)^{\frac{3}{2}})^{\frac{2}{3}}=27^{\frac{2}{3}}\)

Left - hand side: \((11 + x)^{\frac{3}{2}\times\frac{2}{3}}=11 + x\)

Right - hand side: \(27^{\frac{2}{3}}=(3^3)^{\frac{2}{3}}=3^{3\times\frac{2}{3}}=3^2 = 9\)

So \(11+x = 9\)

Subtract \(11\) from both sides: \(x=9 - 11=-2\)

Wait, but let's verify:

If \(x = - 2\), then \(11+x=9\), and \(9^{\frac{3}{2}}=(9^{\frac{1}{2}})^3=3^3 = 27\), which matches the right - hand side of the original equation. So the solution is \(x=-2\)

Wait, maybe I made a mistake in the initial mis - reading, but according to the steps above, the solution for \(x\) is \(-2\)

Answer:

\(x = - 2\)