7 If logs(x) = 2, find x. 8 Solve: 3x+1 = 81 9 A bacterial culture doubles every hour. If there are initially 100 bacteria, how many will there be after 5 hours? 10 The pH of a solution is given by pH = -log10[H+]. If [H+] = 10-7, what is the pH?

: If $\log_5(x) = 2$, find $x$. (Assuming the base of the logarithm is 5, as this leads to one of the options.)

Step 1: Convert the logarithmic equation to an exponential equation. $\log_5(x) = 2 \implies x = 5^2$ Step 2: Calculate the value of $x$. $x = 25$ The correct option is D. $\boxed{D. x = 25}$

: Solve: $3^{x+1} = 81$.

Step 1: Express 81 as a power of 3. $81 = 3^4$ Step 2: Equate the exponents since the bases are the same. $3^{x+1} = 3^4 \implies x+1 = 4$ Step 3: Solve for $x$. $x = 4 - 1$ $x = 3$ The correct option is B. $\boxed{B. x = 3}$

: A bacterial culture doubles every hour. If there are initially 100 bacteria, how many will there be after 5 hours?

Step 1: Identify the initial number of bacteria and the growth factor. Initial bacteria $P_0 = 100$. The culture doubles every hour, so the growth factor is 2. The number of hours is $t = 5$. Step 2: Use the exponential growth formula $P(t) = P_0 \cdot 2^t$. $P(5) = 100 \cdot 2^5$ Step 3: Calculate the value. $P(5) = 100 \cdot 32$ $P(5) = 3200$ The correct option is C. $\boxed{C. 3200}$

: The pH of a solution is given by $\text{pH} = -\log_{10}[H^+]$. If $[H^+] = 10^{-7}$, what is the pH?

Step 1: Substitute the given concentration of hydrogen ions into the pH formula. $pH = -\log_{10}(10^{-7})$ Step 2: Use the logarithm property $\log_b(b^x) = x$. $pH = -(-7)$ Step 3: Simplify the expression. $pH = 7$ The correct option is C. $\boxed{C. 7}$

: Simplify: $(2^{-3} \times 2^5) \div 2^2$.

Step 1: Apply the product rule for exponents ($a^m \times a^n = a^{m+n}$) to the terms inside the parenthesis. $2^{-3} \times 2^5 = 2^{-3+5} = 2^2$ Step 2: Substitute this result back into the expression. $2^2 \div 2^2$ Step 3: Apply the quotient rule for exponents ($a^m \div a^n = a^{m-n}$). $2^2 \div 2^2 = 2^{2-2} = 2^0$ Step 4: Simplify the expression. $2^0 = 1$ The correct option is A. $\boxed{A. 1}$

: Find the remainder when 47 is divided by 5.

Step 1: Perform the division of 47 by 5. $47 = 5 \times q + r$ Step 2: Determine the quotient ($q$) and the remainder ($r$). $47 = 5 \times 9 + 2$ The remainder is 2. The correct option is B. $\boxed{B. 2}$

: Solve: $3x \equiv 2 \pmod{5}$.

Step 1: Find the multiplicative inverse of 3 modulo 5. We need a number $k$ such that $3k \equiv 1 \pmod{5}$. $3 \times 2 = 6 \equiv 1 \pmod{5}$ So, the inverse of 3 modulo 5 is 2. Step 2: Multiply both sides of the congruence by the inverse (2). $2 \times (3x) \equiv 2 \times 2 \pmod{5}$ $6x \equiv 4 \pmod{5}$ Step 3: Simplify the congruence. Since $6 \equiv 1 \pmod{5}$. $1x \equiv 4 \pmod{5}$ $x \equiv 4 \pmod{5}$ The correct option is D. $\boxed{D. x \equiv 4 \pmod{5}}$

: What is $23 \pmod{7}$?

Step 1: Divide 23 by 7. $23 = 7 \times q + r$ Step 2: Determine the quotient ($q$) and the remainder ($r$). $23 = 7 \times 3 + 2$ The remainder is 2. The correct option is B. $\boxed{B. 2}$

: Solve: $2x + y = 7$ and $x - y = 2$.

Step 1: Use the elimination method by adding the two equations. $(2x + y) + (x - y) = 7 + 2$ $3x = 9$ Step 2: Solve for $x$. $x = \frac{9}{3}$ $x = 3$ Step 3: Substitute the value of $x$ into the second equation ($x - y = 2$) to find $y$. $3 - y = 2$ $y = 3 - 2$ $y = 1$ The solution is $x=3, y=1$. The correct option is A. $\boxed{A. x=3, y=1}$

: Solve: $3x + 2y = 12$ and $x + y = 5$.

Step 1: From the second equation, express $y$ in terms of $x$. $y = 5 - x$ Step 2: Substitute this expression for $y$ into the first equation. $3x + 2(5 - x) = 12$ Step 3: Solve for $x$. $3x + 10 - 2x = 12$ $x + 10 = 12$ $x = 12 - 10$ $x = 2$ Step 4: Substitute the value of $x$ back into the expression for $y$. $y = 5 - 2$ $y = 3$ The solution is $x=2, y=3$. The correct option is A. $\boxed{A. x=2, y=3}$