Problem 4 asks us to find the least value of $x$ that satisfies the equation $4x \equiv 7 \pmod{9}$. Problem 5 asks us to express $1 + 2\log_{10}3$ in the form $\log_{10}q$.

1. Problem Description

Problem 4 asks us to find the least value of $x$ that satisfies the equation $4x \equiv 7 \pmod{9}$.

Problem 5 asks us to express $1 + 2\log_{10}3$ in the form $\log_{10}q$.

2. Solution Steps

Problem 4:

We want to find the smallest positive integer $x$ such that $4x \equiv 7 \pmod{9}$. We are looking for a multiple of 4 that is 7 more than a multiple of

9. We can rewrite the congruence as $4x = 7 + 9k$ for some integer $k$. We can test values of $k$ to see if $7 + 9k$ is divisible by

4.

If $k=0$, $7 + 9(0) = 7$, not divisible by

4. If $k=1$, $7 + 9(1) = 16$, divisible by

4. $16/4 = 4$.

So, $x = 4$. We can check that $4(4) = 16 \equiv 7 \pmod{9}$. Since $16 - 7 = 9$, the smallest value of $x$ is indeed

4. Alternatively, we can find the modular inverse of 4 modulo

9. We need to find a number $a$ such that $4a \equiv 1 \pmod{9}$. Since $4(7) = 28 = 3(9) + 1$, $4(7) \equiv 1 \pmod{9}$, so the modular inverse of 4 mod 9 is

7. Then, $4x \equiv 7 \pmod{9}$ becomes $7(4x) \equiv 7(7) \pmod{9}$, so $28x \equiv 49 \pmod{9}$. Then $x \equiv 49 \pmod{9}$. Since $49 = 5(9) + 4$, we have $x \equiv 4 \pmod{9}$.

The smallest value is $x = 4$.

Problem 5:

We want to express $1 + 2\log_{10}3$ in the form $\log_{10}q$.

$1 + 2\log_{10}3 = \log_{10}10 + \log_{10}3^2 = \log_{10}10 + \log_{10}9 = \log_{10}(10 \cdot 9) = \log_{10}90$.

We used the following properties of logarithms:

$a \log_b c = \log_b c^a$

$\log_b a + \log_b c = \log_b (ac)$

$\log_b b = 1$

3. Final Answer

Problem 4: D. 4

Problem 5: A. log10 90