Logarithms

Given a = b = c = 10. Then x = log(a × b × c) = log(10 × 10 × 10) = log(1000). Using logarithm property: log(abc) = log(a) + log(b) + log(c) = log(10) + log(10) + log(10). Assuming base 10: log(10) = 1, so x = 1 + 1 + 1 = 3. The answer is 3, not log 100 or 3 ln 10.

To solve this problem, we need to understand the concept of logarithms and how they relate to exponents and exponentiation.

The problem states that the logarithm (base ten) of five hundred times an integer is three. We can write this information as:

log10(500x) = 3

To find the value of x, we need to isolate it on one side of the equation. We can do this by exponentiating both sides of the equation with base 10:

10^(log10(500x)) = 10^3

500x = 1000

Now, we can solve for x by dividing both sides of the equation by 500:

x = 1000/500

x = 2

Therefore, the correct answer is D. 2.

If x, y, z are in GP, then y² = xz, so 2log y = log x + log z. Let a = log x, b = log y, c = log z. Then 2b = a + c. For the sequence 1/(1+a), 1/(1+b), 1/(1+c) to be in HP, their reciprocals (1+a), (1+b), (1+c) must be in AP. Since 2(1+b) = 2 + 2b = 2 + a + c = (1+a) + (1+c), the reciprocals are in AP, so the original sequence is in HP.

The logarithm of 0 is undefined in real numbers. As x approaches 0 from the positive side, log(x) approaches negative infinity because you're asking "what power must I raise the base to, to get 0?" - no real number works.

The logarithm of 1 is always 0 in any base, because any number raised to the power of 0 equals 1. This is a fundamental property of logarithms: log_b(1) = 0 for all bases b > 0, b ≠ 1.

Using base 10: log(1+1) + log(2+3) = log(2) + log(5) = log(2×5) = log(10) = 1. This uses the logarithm property log(a) + log(b) = log(ab). Since 2 × 5 = 10 and log_10(10) = 1, the statement is true.

We need to solve: log(500n) = 3. This means 500n = 10^3 = 1000, so n = 1000/500 = 2. Therefore, the integer is 2. Option A (450), B (8), and C (1500) are incorrect solutions.

If log4(x) = 12, then x = 4^12. We need log2(x/4). x/4 = 4^12 / 4^1 = 4^11. Since 4 = 2^2, 4^11 = (2^2)^11 = 2^22. Therefore, log2(2^22) = 22.

Log_x(1/8) = -3/2 means x^(-3/2) = 1/8, so x^(3/2) = 8. Squaring both sides: x^3 = 64, therefore x = 4^(1/3) = 4. But we must check if x=1/4 works: (1/4)^(-3/2) = (4)^(3/2) = 8, which is correct.

First evaluate log10(1000) = 3, because 10^3 = 1000. The equation becomes 3 = 8 + v. Solving for v: v = 3 - 8 = -5. This problem tests knowledge of logarithms (log base 10 of 1000) combined with basic linear equation solving.

Assuming base e: let log(x) = t, so the equation becomes 2t - 5t + 6 = 0, giving t² - 5t + 6 = 0. This factors as (t-2)(t-3) = 0, so t = 2 or t = 3. Therefore x = e² or x = e³.

The logarithm of 1 in any base (including base 10 for 'log' and base 'e' for 'ln') is always 0. Therefore, 0 + 0 = 0.

Given log27/log12 = a, we know log27 = log(3^3) = 3log3, and log12 = log(3×4) = log3 + log4 = log3 + 2log2. We need log16/log6 = log(2^4)/log(2×3) = 4log2/(log2+log3). Substituting and simplifying using the given ratio yields the answer 4(3-a)/(3+a). This requires logarithmic properties and algebraic manipulation.

The depth of a complete binary tree is the number of levels from root to deepest leaf. For n nodes, the depth equals ⌊log₂n⌋ + 1 because a tree of depth d can hold at most 2^d - 1 nodes. This logarithmic relationship reflects how nodes double at each level in a perfect binary tree.