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Chapter 3: Problem 17

Find the smallest integer \(n\) such that \(7^{n}>10^{100}\).

Short Answer

Expert verified
The smallest integer \(n\) such that \(7^{n}>10^{100}\) is \(n = 62\).

Step by step solution

01

Convert inequality into logarithmic form

To convert the given inequality, \(7^{n}>10^{100}\), into logarithmic form, we can take the logarithm of both sides. In this case, we'll use the natural logarithm (base \(e\)). \[\ln{(7^{n})} > \ln{(10^{100})}\]
02

Apply logarithm properties

We can now apply the power rule of logarithms to the inequality, which states that \(\ln{a^b} = b\ln{a}\). This will allow us to bring the exponents out front. \[n\ln{7} > 100\ln{10} \]
03

Divide by \(\ln{7}\)

To isolate \(n\), we can divide both sides of the inequality by \(\ln{7}\). \[n > \frac{100\ln{10}}{\ln{7}}\]
04

Calculate the value of the expression on the right

Let's compute the value of the expression on the right side of the inequality: \[n > \frac{100\ln{10}}{\ln{7}} \approx 61.358\]
05

Find the smallest integer value for n

Since \(n\) must be an integer, the smallest possible value for \(n\) that satisfies the inequality is 62 (as 61 does not satisfy the inequality). So, the smallest integer \(n\) such that \(7^{n}>10^{100}\) is \(n = 62\).

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