Question

1. find n.

a. 5^{11}cdot5^{3}=5^{n}
b. x^{15}cdot x^{n}=x^{9}
c. 6^{17}cdot6^{n}=6^{-2}
d. \frac{8^{7}}{8^{3}} = 8^{n}
e. \frac{x^{9}}{x^{n}}=x^{12}
f. 4^{n}=4^{5}=1

Explanation:

Step1: Use the rule of exponents \(a^m\cdot a^n=a^{m + n}\)

a.

For \(5^{11}\cdot5^{3}=5^{n}\), by the rule \(a^m\cdot a^n=a^{m + n}\), we have \(5^{11 + 3}=5^{n}\), so \(5^{14}=5^{n}\), then \(n = 14\).

b.

For \(x^{15}\cdot x^{n}=x^{9}\), using \(a^m\cdot a^n=a^{m + n}\), we get \(x^{15 + n}=x^{9}\). Then \(15 + n=9\), and \(n=9 - 15=- 6\).

c.

For \(6^{17}\cdot6^{n}=6^{-2}\), by \(a^m\cdot a^n=a^{m + n}\), we have \(6^{17 + n}=6^{-2}\). So \(17 + n=-2\), and \(n=-2 - 17=-19\).

d.

For \(\frac{8^{7}}{8^{3}}=8^{n}\), use the rule \(\frac{a^m}{a^n}=a^{m - n}\), then \(8^{7-3}=8^{n}\), so \(8^{4}=8^{n}\), and \(n = 4\).

e.

For \(\frac{x^{9}}{x^{n}}=x^{12}\), by \(\frac{a^m}{a^n}=a^{m - n}\), we have \(x^{9 - n}=x^{12}\). So \(9 - n=12\), and \(n=9 - 12=-3\).

f.

For \(\frac{4^{n}}{4^{5}}=1\), since \(1 = 4^{0}\) and \(\frac{a^m}{a^n}=a^{m - n}\), we have \(4^{n-5}=4^{0}\). Then \(n-5 = 0\), and \(n = 5\).

Answer:

a. \(n = 14\)
b. \(n=-6\)
c. \(n=-19\)
d. \(n = 4\)
e. \(n=-3\)
f. \(n = 5\)