Solve the inequality (3 x 10^9)/4^x < 1, and give the smallest integer value of x that satisfies it.

Step-by-step solution

1. 1

   Clear the denominator safely. An exponential $4^x$ is strictly positive for every real $x$, so multiplying both sides by it cannot flip the inequality:

   $3\times10^9<4^x,\qquad\text{i.e.}\qquad 4^x>3\times10^9$

   This is the whole reason no case analysis is needed here, unlike inequalities with a polynomial denominator.

2. 2

   Take logarithms base 4. Because $4>1$, the function $\log_4$ is increasing, so it preserves the direction:

   $x>\log_4\!\left(3\times10^9\right)$

3. 3

   Convert with the change-of-base formula. Calculators do not have $\log_4$, so use

   $\log_4 N=\frac{\log_{10}N}{\log_{10}4}$

   For the numerator, split the product with the log rules:

   $\log_{10}\!\left(3\times10^9\right)=\log_{10}3+\log_{10}10^9=0.4771+9=9.4771$

4. 4

   Divide by log 4. With $\log_{10}4=0.6021$:

   $x>\frac{9.4771}{0.6021}=15.7412\ldots$

   So the real solution set is $x>15.7412$, or exactly $x>\log_4\!\left(3\times10^9\right)$.

5. 5

   Read off the smallest integer. The least integer strictly greater than $15.7412$ is

   $x=16$

   A quick estimate confirms the size: $4^{15}=2^{30}\approx1.07\times10^9$, which is less than $3\times10^9$, while $4^{16}=2^{32}\approx4.29\times10^9$, which is more.

6. 6

   Verify both sides of the boundary. At $x=16$: $\dfrac{3\times10^9}{4^{16}}=\dfrac{3\times10^9}{4.295\times10^9}=0.6985<1$ ✓. At $x=15$: $\dfrac{3\times10^9}{1.074\times10^9}=2.795>1$ ✗. And at the exact boundary $x=15.7412$, the quotient equals $1.000000$ ✓.