SCSA Mathematics Methods The logarithmic function

This value is incorrect. It is close to $10^{-0.5}$ rather than the required $10^{-0.2}$.

This value is incorrect and does not result from solving the logarithmic equation for $[\text{H}^+]$.

This result comes from calculating $10^{0.2}$, which incorrectly ignores the negative sign in the pH formula.

Incorrect. While this point lies on the graph of $f(x)$ (since $x \approx \ln(5)$ gives $f(x) = 15$), $g(1.609) \approx 0.6$. The intersection point must satisfy both equations.

Incorrect. This point lies on the graph of $g(x)$ (since $x \approx \ln(3)$ gives $g(x) = 1$), but $f(1.099) \approx 11$. The functions are not equal at this x-value.

Incorrect. This point lies on the graph of $g(x)$ (since $x \approx \ln(1.5)$ gives $g(x) = 2$), but $f(0.4065) \approx 8$. The y-values differ significantly.

This is incorrect. The expression evaluates to $x$, not the constant 0. Furthermore, $\ln(x)$ is undefined at $x=0$.

This is incorrect. The expression would only equal 1 if $x=1$ (since $\ln(1)=0$ and $e^0=1$), but it is not equal to 1 for all values of $x$.

This is incorrect. This result corresponds to $e^{-\ln(x)} = e^{\ln(1/x)} = \frac{1}{x}$, not the original expression.

This answer is incorrect. It corresponds to evaluating an antiderivative like $\frac{1}{2}\ln(2x)$ only at the upper limit $x=3$, neglecting to subtract the value at the lower limit.

This is incorrect. The evaluation of the definite integral depends on the limits $1$ and $3$, which results in $\ln(3)$, not $\ln(5)$.

This option is incorrect. It may result from an arithmetic error or misapplying logarithmic properties, as the integral yields $\ln(3) - \ln(1)$, not $\ln(4)$.

This is the value of the function $N(3) = 15 \ln(22) \approx 46$. This represents the number of koalas (or the population increase) at year 3, rather than the rate at which the population is changing.

This value appears to be the result of calculating $N(3) - 20 \approx 26$. This subtracts the initial population from the model's value at $t=3$, which does not represent the instantaneous rate of change.

This is an incorrect value. It does not correspond to the derivative at $t=3$ or the function value, likely resulting from a calculation error.

This value represents the acceleration at $t=4$, found by differentiating the velocity function: $a(4) = v'(4) = \frac{2}{4+1} = 0.40 \text{ m/s}^2$.

This value corresponds to $\ln(5)$, which is an intermediate value in the calculation but not the final integral result.

This is the instantaneous velocity of the marble at $t=4$, calculated as $v(4) = 2\ln(5) \approx 3.22 \text{ m/s}$, rather than the total distance traveled.

If $n = 2^k$, then $x = \log_2(2^k + 1)$, which does not result in an integer value for $x$.

If $n = 2^{k-1}$, then $x = \log_2(2^{k-1} + 1)$, which does not evaluate to an integer for positive integers $k$.

If $n = 2k - 1$, then $x = \log_2(2k)$, which is only an integer when $k$ itself is a power of 2, not for all positive integers $k$.

If $n = 2k$, then $x = \log_2(2k + 1)$, which does not yield an integer for positive integers $k$.

This result comes from incorrectly multiplying the given logarithm by 100 ($0.778 \times 100$). However, $\log_{10}(100 \times 6) = \log_{10} 6 + \log_{10} 100$, which involves addition, not multiplication.

This option incorrectly adds 10 to the given value. Since $600 = 6 \times 10^2$, you must add the exponent 2 (because $\log_{10} 100 = 2$), not the base 10.

This value represents $2 \times \log_{10} 6$, which equals $\log_{10}(6^2) = \log_{10} 36$. The logarithm of a product ($6 \times 100$) requires adding the logs, not multiplying the log value by 2.