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For a geometric sequence the nth term is u_n = u_1 * r^(n-1). Here u_6 = 3 * 2^5 = 3 * 32 = 96. The exponent is n-1, not n, because the first term already counts as one multiplication by zero factors of r.
The nth term of an arithmetic sequence is u_n = u_1 + (n-1)d. Substituting gives u_20 = 7 + 19 * (-3) = 7 - 57 = -50. The multiplier is 19 because there are only 19 steps of d between the first and twentieth terms.
This is an arithmetic series with u_1 = 2, d = 2 and n = 40, so u_40 = 80. Using S_n = n(u_1 + u_n)/2 gives S_40 = 40(2 + 80)/2 = 40 * 41 = 1640. Equivalently the sum equals n(n+1) for the first n even integers.
A geometric series converges only when |r| < 1, and then S_infinity = u_1/(1 - r). Setting 8/(1 - r) = 24 gives 1 - r = 8/24 = 1/3, so r = 2/3. This value satisfies |r| < 1, so the sum to infinity genuinely exists.
Use S_n = u_1(r^n - 1)/(r - 1) = 5(3^n - 1)/2. Requiring S_n > 10000 gives 3^n - 1 > 4000, so 3^n > 4001. Since 3^7 = 2187 and 3^8 = 6561, the smallest integer n is 8.
The nth term is 4 + 7(n-1). Setting 4 + 7(n-1) = 95 gives 7(n-1) = 91, so n-1 = 13 and n = 14. Checking: the 14th term is 4 + 7 * 13 = 95, as required.
Here u_1 = 2 and r = 3; solving 2 * 3^(n-1) = 1458 gives 3^(n-1) = 729 = 3^6, so n = 7. Then S_7 = 2(3^7 - 1)/(3 - 1) = 2(2187 - 1)/2 = 2186. The sum is one less than the next term 2187 because the first term is 2.
Repeated percentage growth is modelled by a geometric sequence with multiplier 1 + 4/100 = 1.04 per year. After 10 years the value is 50000 * 1.04^10, which is compound growth rather than simple growth. The exponent counts the number of yearly multiplications.
Depreciation by 12% per year leaves 88% of the value each year, so the multiplier is 0.88. The value after 3 years is 20000 * 0.88^3 = 20000 * 0.681472 = 13629.44, which rounds to 13629 dollars. The exponent 3 counts three yearly multiplications.
Scientific notation writes a number as a * 10^k with 1 <= |a| < 10 and k an integer. Here 1500000 = 1.5 * 1000000 = 1.5 * 10^6, and 1.5 lies in the required range. The exponent 6 counts the six places the decimal point moves left.
Rewrite both numbers with the same power of ten: 3.2 * 10^4 + 0.45 * 10^4 = 3.65 * 10^4. Adding requires matching exponents before the coefficients can be combined. The result 36500 is already in correct scientific form as 3.65 * 10^4.
Divide the coefficients and subtract the exponents: (8.4 / 2.1) * 10^(7-3) = 4 * 10^4. Since 4 lies between 1 and 10, this is already in scientific notation. Subtracting exponents follows from 10^7 / 10^3 = 10^4.
Since 8 = 2^3, the power rule gives log 8 = 3 log 2 = 3 * 0.301 = 0.903. The power rule log(a^n) = n log a converts the exponent into a multiplier. This places log 8 close to log 10 = 1, which is consistent.
Write 125 as 5^3 so both sides have the same base: 5^(2x-1) = 5^3. Equating exponents gives 2x - 1 = 3, so 2x = 4 and x = 2. The base-5 exponential function is one-to-one, so equal powers of 5 must have equal exponents.
Rewrite the logarithmic equation in exponential form: x + 3 = 2^4 = 16. Then x = 16 - 3 = 13. Substituting back gives log base 2 of 16 = 4, which confirms the solution.
Apply the power to both factors: (2 * 10^3)^3 = 2^3 * (10^3)^3 = 8 * 10^9. The power rule for exponents multiplies the inner exponent by the outer one, so 3 * 3 = 9. The coefficient 8 is within the range 1 to 10.
For compound interest the amount is A = P(1 + r/k)^(kt), where r is the nominal annual rate and k the number of compounds per year. With monthly compounding k = 12, giving 12000 * (1 + 0.06/12)^(12*5). Both the periodic rate and the number of periods must reflect monthly compounding.
Subtracting the term equations gives u_8 - u_3 = 5d = 31 - 11 = 20, so d = 4. Then u_3 = u_1 + 2d = u_1 + 8 = 11, giving u_1 = 3. The gap between the 3rd and 8th terms is five common differences.
Dividing gives u_5 / u_2 = r^3 = 48 / 6 = 8, so r = cube root of 8 = 2. There are three ratios between the 2nd and 5th terms, hence the cube root. Checking: u_2 = 6 means u_1 = 3, and the 5th term is 3 * 2^4 = 48.
Multiplying powers with the same base adds exponents: x^5 * x^-2 = x^3. Dividing then subtracts exponents: x^3 / x^4 = x^(3-4) = x^-1. This equals 1/x, which is a valid simplified form.
The arithmetic sum formula is S_n = n(u_1 + u_n)/2. Substituting gives S_30 = 30(5 + 92)/2 = 15 * 97 = 1455. The factor n/2 = 15 is the average of the first and last terms multiplied by the number of terms.
The series converges because |-1/2| < 1, so S_infinity = u_1/(1 - r) = 12/(1 - (-1/2)) = 12/(3/2) = 8. The negative ratio is handled by the subtraction of a negative, which increases the denominator to 3/2.
Model the value as V = V_0 * 2^(t/9). Setting 2^(t/9) = 5 gives t/9 = log 5 / log 2, so t = 9 * log 5 / log 2 = 9 * 2.322 = 20.9 years. Rounding gives 21 years, which lies between two and three doubling periods.
In an exponential model y = a * b^x with a > 0, the behaviour depends on the base b. If 0 < b < 1 the function decreases as x increases, which is exponential decay; here b = 0.7. The constant 100 affects only the starting value, not the growth or decay pattern.
The quotient rule gives log 54 - log 2 = log(54/2) = log base 3 of 27. Since 27 = 3^3, the value is 3. Subtracting logarithms with the same base corresponds to dividing the arguments.
Apply the recurrence step by step: u_2 = 2(3) + 1 = 7, then u_3 = 2(7) + 1 = 15, then u_4 = 2(15) + 1 = 31. Each term depends on the immediately preceding term, so the recursion must be applied three times starting from u_1 = 3.
Each term is double the previous one, so the sequence is geometric with first term 5 and common ratio 2. The general term of a geometric sequence is u_n = u_1 r^(n-1), giving u_n = 5(2)^(n-1), which correctly returns 5 when n = 1.
The general term is u_n = 7 + 4(n-1). Setting 7 + 4(n-1) > 200 gives 4(n-1) > 193, so n-1 > 48.25 and hence n > 49.25. The smallest integer satisfying this is n = 50, and checking gives u_50 = 203 while u_49 = 199.
Because |r| = 1/4 < 1, the series converges and the sum to infinity is S = u_1/(1 - r). Substituting gives 80/(1 - 1/4) = 80/(3/4) = 320/3, which is approximately 106.7.
The quotient law of logarithms states that log_a(m) - log_a(n) = log_a(m/n), so the expression equals log_5(200/8) = log_5(25). Since 5^2 = 25, this simplifies to 2.
Write both sides with base 2: 4^(x+1) = 2^(2x+2) and 8^(x-1) = 2^(3x-3). Equating exponents gives 2x + 2 = 3x - 3, so x = 5. Checking: 4^6 = 4096 and 8^4 = 4096, confirming the solution.
The quarterly rate is 0.048/4 = 0.012, so the growth factor per quarter is 1.012. Solving 1.012^n = 1.6 gives n = log(1.6)/log(1.012) ≈ 39.4 quarters, which is about 9.85 years and rounds to 9.75 years at the nearest quarter-year.
Substituting x = 4 gives f(4) = 3(4) − 5 = 12 − 5 = 7. The order of operations means multiplication is performed before subtraction.
g(−3) = (−3)² − 2(−3) = 9 + 6 = 15. Squaring a negative gives a positive, and subtracting a negative product adds.
Adding 3 outside the function translates the whole graph vertically upwards by 3 units. The y-intercept therefore moves from (0, 4) to (0, 7).
A constant function maps every input to the same output, so it fails the horizontal line test and is not one-to-one. An inverse function requires distinct inputs to give distinct outputs, so h has no inverse.
Substituting x = 3 and f(x) = 11 gives 11 = 2(3) + k, so 11 = 6 + k and k = 5.
The expression under a square root must be non-negative, so x − 4 ≥ 0, giving x ≥ 4. The value x = 4 is allowed because √0 = 0 is defined.
Since x² ≥ 0 for all real x, adding 1 gives f(x) ≥ 1, and the value 1 occurs at x = 0. So the range is y ≥ 1.
Set y = 4x − 7 and solve for x: x = (y + 7)/4. Then f⁻¹(9) = (9 + 7)/4 = 16/4 = 4.
Replacing x by x − 2 shifts the graph horizontally by 2 units in the positive x-direction, that is, to the right. A change inside the function affects the x-coordinates in the opposite sense to the sign shown.
The composite (f ∘ g)(x) means f(g(x)). Here g(3) = 9, then f(9) = 2(9) + 1 = 19.
The gradient is (13 − 5)/(6 − 2) = 8/4 = 2. Using the point (2, 5): 5 = 2(2) + c gives c = 1, so y = 2x + 1.
Completing the square gives x² − 6x + 5 = (x − 3)² − 9 + 5 = (x − 3)² − 4. The vertex is therefore at (h, k) = (3, −4).
The denominator is zero at x = 3, giving a vertical asymptote there. For large |x| the ratio of leading coefficients is 1/1, so the horizontal asymptote is y = 1.
For a base greater than 1 the exponential function is increasing. Powers of 2 are always positive but never reach 0, so the range is y > 0.
The x-coordinate is unchanged by a vertical stretch and shift. The y-coordinate becomes 2(6) − 1 = 11, so the image point is (1, 11).
(g ∘ f)(x) = g(f(x)) = g(x + 3) = 2(x + 3) − 1 = 2x + 6 − 1 = 2x + 5.
Write y = x² − 4x = (x − 2)² − 4, so (x − 2)² = y + 4 and x = 2 ± √(y + 4). Since the domain restricts x ≥ 2, the positive root is taken, giving f⁻¹(x) = 2 + √(x + 4).
Vertex form gives f(x) = (x − 2)² − 1 = x² − 4x + 3, so b = −4 and c = 3. Equivalently, the axis of symmetry x = −b/2 = 2 gives b = −4, then f(2) = 4 − 8 + c = −1 gives c = 3.