TMUA Exponentials & logs questions

Answer: C · Method: Use the laws and definition of logarithms

1. Let $x = 2^t$, so $\log_2 x = t$. Since $4^{t/2} = 2^t$ and $8^{t/3} = 2^t$, $\log_4 x = \tfrac{t}{2}$ and $\log_8 x = \tfrac{t}{3}$.
2. So $t + \tfrac{t}{2} + \tfrac{t}{3} = \tfrac{11}{6}t = 11$, giving $t = 6$ and $x = 2^6 = 64$.

Answer: C · Method: Solve an exponential equation as a quadratic

1. Let $u = 2^x$, so $4^x = u^2$ and $u^2 - 6u + 8 = 0$, giving $u = 2$ or $u = 4$.
2. So $x = 1$ or $x = 2$, and the sum is $3$.

Why the tempting wrong answers are wrong

• D: $6$ is the sum of the values of $u = 2^x$, not of $x$.

Answer: C · Method: Use the laws and definition of logarithms

1. Let $X = \log_2 x$ and $Y = \log_2 y$. Then $X + Y = 5$ and $X - Y = 1$.
2. Adding gives $X = 3$, so $Y = 2$. Then $x = 2^3 = 8$ and $y = 2^2 = 4$.
3. So $x + y = 12$.

Why the tempting wrong answers are wrong

• A: $6$ is $XY = \log_2 x \times \log_2 y$. Turn the logs back into $x$ and $y$ first.
• F: $32$ is $xy = 2^{X + Y}$, not $x + y$.

Answer: C · Method: Solve an exponential equation as a quadratic

1. Let $u = 2^x$. Then $4^x = u^2$, so $u^2 + 1u - 2 = 0$, giving $u = 1$ or $u = -2$.
2. $2^x = 1$ gives $x = 0$. $2^x = -2$ is impossible, since $2^x > 0$. The sum is $0$.

Why the tempting wrong answers are wrong

• B: That's the sum of the values of $2^x$, not of $x$.
• D: $2^x$ is always positive, so $2^x = -2$ has no solution.

Answer: C · Method: Combine logarithms, then check the domain

1. Combine the logs: $\log_{2}\big((x + 4)(x - 3)\big) = 3$, so $(x + 4)(x - 3) = 8$.
2. Expanding: $x^2 + x - 20 = 0$, with roots $x = 4$ and $x = -5$.
3. Both brackets must be positive. At $x = -5$, $x - 3 = -8 < 0$, so only $x = 4$ works.

Why the tempting wrong answers are wrong

• A: At $x = -5$, $x - 3$ is negative, so the logarithm isn't defined.
• B: One of the roots makes a logarithm of a negative number. Always check the domain.

Answer: D · Method: Combine logarithms, then check the domain

1. Combine the logs: $\log_{2}\big((x - 2)(x - 5)\big) = 2$, so $(x - 2)(x - 5) = 4$.
2. Expanding: $x^2 - 7x + 6 = 0$, with roots $x = 6$ and $x = 1$.
3. Both brackets must be positive. At $x = 1$, $x - 5 = -4 < 0$, so only $x = 6$ works.

Why the tempting wrong answers are wrong

• A: At $x = 1$, $x - 5$ is negative, so the logarithm isn't defined.
• E: One of the roots makes a logarithm of a negative number. Always check the domain.

Answer: C · Method: Substitute to get a quadratic, then use the substitute's range

1. Let $u = 3^{\cos x}$. Since $-1 \le \cos x \le 1$, $u$ takes every value in $\tfrac{1}{3} \le u \le 3$.
2. The expression is $u^2 - 6u = (u - 3)^2 - 9$, which is decreasing for $u \le 3$.
3. So the least value is $-9$ (at $u = 3$) and the greatest is $\left(\tfrac{1}{3} - 3\right)^2 - 9 = \tfrac{64}{9} - 9 = -\tfrac{17}{9}$ (at $u = \tfrac{1}{3}$). The difference is $-\tfrac{17}{9} + 9 = \tfrac{64}{9}$.

Why the tempting wrong answers are wrong

• A: $\tfrac{17}{9}$ is the size of the greatest value, not the difference between the greatest and least values.
• E: $3^{\cos x}$ is never $0$: it always lies between $\tfrac{1}{3}$ and $3$. Always find the range of a substitution.