The smallest integer $n$ for which $4^n > 17^{19}$ holds, is closest to

We need to find the smallest integer $n$ such that $4^n > 17^{19}$.

Comparing these huge numbers directly is impossible! We need a smarter approach.

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When we have equations or inequalities with different bases and exponents, logarithms are our best friend. They convert multiplication into addition and powers into multiplication.

If $a > b$ (where $a, b > 0$), then $\log(a) > \log(b)$.

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Starting with: $4^n > 17^{19}$

Take the natural logarithm of both sides:

$\ln(4^n) > \ln(17^{19})$

Use the logarithm property $\ln(a^b) = b \ln(a)$:

$n \ln(4) > 19 \ln(17)$

Solve for $n$:

$n > \dfrac{19 \ln(17)}{\ln(4)}$

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We need to find $\ln(4)$ and $\ln(17)$:

For $\ln(4)$: Since $4 = 2^2$, we have $\ln(4) = \ln(2^2) = 2\ln(2) \approx 2 \times 0.693 = 1.386$

For $\ln(17)$: $\ln(17) \approx 2.833$

Substituting:

$n > \dfrac{19 \times 2.833}{1.386} = \dfrac{53.827}{1.386} \approx 38.84$

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Since we need the smallest integer $n$ such that $4^n > 17^{19}$, and we found $n > 38.84$, the answer is:

$n = 39$

For $n = 38$: $4^{38}$ would be slightly smaller than $17^{19}$

For $n = 39$: $4^{39}$ is the first power of 4 that exceeds $17^{19}$

When comparing exponential expressions with different bases, logarithms transform the problem into simple arithmetic!

Therefore, the smallest integer $n$ is 39.