Exponential and logarithmic functions (Mathematics II)

A negative exponent means the reciprocal, so 2⁻³=1/2³=1/8. The option −8 misreads it as simply attaching a minus sign, 1/6 computes 2×3, and −1/8 mixes both errors.

For a≠0 we have a⁰=1, because 5²÷5²=5⁰ and the actual value is 25÷25=1. The option 0 is the classic misunderstanding that a zeroth power must be zero, and 1/5 confuses it with 5⁻¹.

8^(2/3)=(the cube root of 8)²=2²=4. The option 2 stops at 8^(1/3), 16 treats the exponent as a multiplication like 8×2, and 16/3 computes 8×2/3.

The exponents give 3+5−2=6, so the answer is a⁶. The option a¹⁰ adds even for the division, as 3+5+2, a⁸ forgets ÷a² and stops at 3+5, and a⁴ computes (3+5)÷2.

Since (2³)²=2⁶, we get 2⁶×2⁻⁴=2²=4. The option 2 takes the exponent of (2³)² as 3+2=5 and computes 2⁵⁻⁴, 64 forgets to multiply by 2⁻⁴ and leaves 2⁶, and 1/4 mishandles the sign and gives 2⁻².

A product of radicals of the same index is the radical of the product: the cube root of 16×4=64 is 4. The option 8 confuses this with the square root of 64, the cube root of 20 comes from adding 16+4, and 2 confuses it with the cube root of 8.

Writing everything to base 2 gives √2=2^(1/2), the cube root of 4 as 2^(2/3), and the fourth root of 8 as 2^(3/4). Since the base 2 is greater than 1, the order of the exponents 1/2<2/3<3/4 is exactly the order of the values, so √2 comes first, then the cube root of 4, then the fourth root of 8.

A negative exponent means the reciprocal, so (1/2)⁻²=2²=4. The option 1/4 ignores the sign and computes (1/2)², while −4 and −1/4 come from the misunderstanding that a negative exponent produces a negative number.

Since 3⁰=1 the graph always passes through (0, 1), and because the base 3 is greater than 1 it increases. Since y>0 always, it never meets the x-axis, which is an asymptote. The graph decreases when the base satisfies 0<a<1.

4ˣ+4⁻ˣ=(2ˣ+2⁻ˣ)²−2·2ˣ·2⁻ˣ=3²−2=7, using 2ˣ·2⁻ˣ=2⁰=1. The option 9 forgets the −2, 11 has the sign error +2, and 6 comes from computing 2×3.

It answers the question of what power of 2 gives 8, and since 2³=8 the value is 3. The option 4 divides as 8÷2, 16 multiplies as 2×8, and 2 simply repeats the base.

Since 3⁰=1, we have log_3 1=0. For any base, log_a 1=0. The option 1 confuses this with log_a a=1, 1/3 is the value of 3⁻¹, and 3 simply repeats the base.

log_2 8=3 and log_2 4=2, so the sum is 3+2=5, which also equals log_2 32 for the product of the arguments. The option 6 multiplies as 3×2, 12 adds the arguments as 8+4, and 1 treats it as the quotient log_2(8/4).

By the product rule, log_10(5×2)=log_10 10=1. The option log_10 7 adds the arguments as 5+2, log_10 3 subtracts them as 5−2, and 10 drops the logarithm and reports only the product of the arguments.

Since √8=8^(1/2), we get log_2 √8=(1/2)log_2 8=(1/2)×3=3/2. The option 3 ignores the radical and gives log_2 8, 1/2 reports only the exponent, and 2√2 is the value of √8 itself.

By the quotient rule, log_3(18/2)=log_3 9=2, since 3²=9. The option 9 drops the logarithm and reports only 18÷2, 16 subtracts the arguments as 18−2, and 1 divides the argument too far, turning log_3(18/2) into log_3 3, whose value is 1.

By the change of base formula, log_4 8=log_2 8/log_2 4=3/2. The option 2/3 inverts numerator and denominator, 2 divides as 8÷4, and 1/2 divides as 4÷8.

Since 45=3²×5, we get log_2 45=2log_2 3+log_2 5=2a+b. The option a+2b splits 45 wrongly as 3×5², a+b drops the square as though 45 were 15, and 2ab turns a coefficient into a multiplication.

By change of base, log_3 4=log_2 4/log_2 3, so the product is log_2 4=2. The option 1 confuses this with log_2 3×log_3 2=1, a reciprocal pair, while log_2 12 and 12 come from wrongly thinking that a product multiplies the arguments.

Since 16=2⁴, comparing exponents gives x=4. The option 8 computes 16÷2, 32 computes 16×2, and 2 confuses it with √16. The basic move is to write both sides as powers of the same base.

Since 1/9=1/3²=3⁻², we get x=−2. The option 2 forgets the minus sign that comes with the reciprocal, −3 confuses it with 1/27=3⁻³, and 3 confuses it with 27. Use 1/aⁿ=a⁻ⁿ to match the bases.

Writing everything to base 2 gives 2²ˣ=2³, so 2x=3 and x=3/2. The option 2/3 solves 2x=3 upside down, 2 comes from assuming 4×2=8, and 4 subtracts as 8−4.

By the definition, x=2⁵=32. The option 25 swaps base and exponent as 5², 10 computes 2×5, and 7 computes 2+5. Everything rests on the restatement log_a x=p ⇔ x=aᵖ.

Under the condition x−1>0 for the argument, x−1=3²=9, so x=10, and indeed x−1=9>0 is acceptable. The option 9 forgets to add 1, 7 adds 1 to 3×2=6, and 4 misreads the power as 3¹=3 and adds 1.

Since 1/8=(1/2)³, the inequality is (1/2)ˣ<(1/2)³. The base 1/2 is less than 1, so the function is decreasing and the inequality sign reverses, giving x>3. The option x<3 is the classic slip of forgetting the reversal. Checking, (1/2)⁴=1/16<1/8.

Note that −2=log_(1/3) 9, since (1/3)⁻²=9. Because the base 1/3 is less than 1, the inequality sign reverses to give x<9, and together with the argument condition x>0 we get 0<x<9. The option x>9 forgets the reversal, and 1/9 comes from the sign error (1/3)².

Putting 2ˣ=t with t>0 gives t²−3t−4=0, that is (t−4)(t+1)=0. Since t>0, only t=4 survives, and 2ˣ=4 gives x=2. The option x=2, −1 forgets to discard t=−1, x=4 reports the value of t as the solution, and x=1 fails on substitution, since 4−6−4=−6≠0.

log_10 2¹⁰=10×0.3010=3.010. From 3≦log_10 2¹⁰<4 we get 10³≦2¹⁰<10⁴, so it has 4 digits; indeed 2¹⁰=1024. The option 3 digits reports the integer part without adding 1, and 10 digits confuses it with the exponent.

log_10 2⁵⁰=50×0.3010=15.05. From 15≦log_10 2⁵⁰<16 we get 10¹⁵≦2⁵⁰<10¹⁶, so it has 16 digits. The option 15 digits forgets to add 1 to the integer part, 50 digits confuses it with the exponent, and 17 digits counts one too many.

Taking common logarithms of both sides gives −0.3010n<−3, so n>3/0.3010=9.96 and a little more. The smallest natural number satisfying this is n=10. Checking, (1/2)⁹≒0.00195 is larger than 0.001, and (1/2)¹⁰≒0.00098 is the first to fall below it. The option 9 rounds 9.96 down, and 3 simply reports the exponent in 0.001=10⁻³.