Question

tell whether ( n = 3 ) makes each equation true. (\frac{8^5}{8^n}=8^2) (marked as true) (\frac{(12^2)^4}{12^3}=\frac{12^n}{12^2}) with options true and false

Explanation:

Step1: Simplify left - hand side of the second equation

First, use the power of a power rule \((a^{m})^{n}=a^{mn}\) on the numerator of \(\frac{(12^{2})^{4}}{12^{3}}\). For \((12^{2})^{4}\), we have \(m = 2\) and \(n=4\), so \((12^{2})^{4}=12^{2\times4}=12^{8}\).
Then, use the quotient rule of exponents \(\frac{a^{m}}{a^{n}}=a^{m - n}\). So \(\frac{12^{8}}{12^{3}}=12^{8-3}=12^{5}\).

Step2: Substitute \(n = 3\) into the right - hand side

The right - hand side of the equation is \(12^{n}\). When \(n = 3\), the right - hand side is \(12^{3}\).
Since \(12^{5}
eq12^{3}\) (because the exponents \(5
eq3\) and for the exponential function \(y = 12^{x}\), when the bases are the same, the values are equal if and only if the exponents are equal), when \(n = 3\), the equation \(\frac{(12^{2})^{4}}{12^{3}}=\frac{12^{n}}{12^{2}}\) (wait, there is a typo in the original equation? Wait, the original equation is \(\frac{(12^{2})^{4}}{12^{3}}=\frac{12^{n}}{12^{2}}\)? Wait, no, let's re - check. Wait, the user wrote \(\frac{(12^{2})^{4}}{12^{3}}=\frac{12^{n}}{12^{2}}\)? Wait, maybe it's a formatting error. Wait, the original equation is \(\frac{(12^{2})^{4}}{12^{3}} = 12^{n}\)? No, the user's equation is \(\frac{(12^{2})^{4}}{12^{3}}=\frac{12^{n}}{12^{2}}\)? Wait, let's re - derive.

Wait, let's re - express the equation correctly. Let's assume the equation is \(\frac{(12^{2})^{4}}{12^{3}}=\frac{12^{n}}{12^{2}}\). Then, cross - multiply: \((12^{2})^{4}\times12^{2}=12^{n}\times12^{3}\). Using the product rule \(a^{m}\times a^{n}=a^{m + n}\) and power of a power rule \((a^{m})^{n}=a^{mn}\), we have \(12^{4\times2+2}=12^{n + 3}\), so \(12^{10}=12^{n + 3}\), then \(n+3 = 10\), \(n = 7\). But if the equation is \(\frac{(12^{2})^{4}}{12^{3}}=12^{n}\), then as we calculated before, \(\frac{(12^{2})^{4}}{12^{3}}=12^{8-3}=12^{5}\), so \(n = 5\).

Wait, maybe the original equation is \(\frac{(12^{2})^{4}}{12^{3}}=\frac{12^{n}}{12^{2}}\) is a mistake, and it should be \(\frac{(12^{2})^{4}}{12^{3}}=12^{n}\)? No, let's check the exponents again. Wait, the left - hand side: \((12^{2})^{4}=12^{8}\), divided by \(12^{3}\) is \(12^{5}\). The right - hand side: \(\frac{12^{n}}{12^{2}}=12^{n - 2}\). So we have the equation \(12^{5}=12^{n - 2}\). Then, by the property of exponents (if \(a^{m}=a^{n}\), then \(m = n\) for \(a>0,a
eq1\)), we get \(5=n - 2\), so \(n=7\). But we are checking for \(n = 3\). So when \(n = 3\), the right - hand side is \(12^{3-2}=12^{1}\), and the left - hand side is \(12^{5}\), so they are not equal.

Alternatively, if the equation is \(\frac{(12^{2})^{4}}{12^{3}}=12^{n}\), then as we calculated, left - hand side is \(12^{5}\), so \(n = 5\). If \(n = 3\), then \(12^{5}
eq12^{3}\). So in either case, when \(n = 3\), the equation is false.

Answer:

false