Exponential and Logarithmic Equations - Pre Assessment

To solve 16^(5x) = 64^(x+7), we express both sides with the same base. 16 = 4^2 and 64 = 4^3, leading to 4^(2(5x)) = 4^(3(x+7)). Alternatively, using base 2 gives 2^(4(5x)) = 2^(6(x+7)). Both setups are correct.

To solve 9(e^x) - 31 = 23, first add 31 to both sides: 9(e^x) = 54. Then divide by 9: e^x = 6. Taking the natural logarithm gives x = ln(6) ≈ 1.79, which matches the correct answer.

To solve 2(16^(x+6)) - 8 = 35, first add 8 to both sides: 2(16^(x+6)) = 43. Then divide by 2: 16^(x+6) = 21.5. Taking log base 16 gives x+6 = log16(21.5). Solving yields x = -4.89, the correct answer.

Combine the logs: log_9(x(x+2)) = log_9(35). This gives x(x+2) = 35. Solving the quadratic x^2 + 2x - 35 = 0, we find x = 5 or x = -7. Only x = 5 is valid since logs of negative numbers are undefined.

Using the properties of logarithms, we can combine the left side: log((x - 1)/(x - 2)) = log 5. This implies (x - 1)/(x - 2) = 5. Solving gives x - 1 = 5(x - 2), leading to x = 9/4, which is the correct answer.

To solve the equation, we need to express 8 as a power of 2. Since 8 = 2^3, we rewrite 8 using base 2. Thus, the correct base to rewrite 8 is 2.

To rewrite the exponential equation 2^(3x+1) = 12 as a logarithm, we use the definition of logarithms: log_b(a) = c means b^c = a. Here, log_2(12) = 3x + 1 is the correct transformation.

To solve log₂(x + 3) = 5, rewrite it in exponential form: x + 3 = 2^5. This simplifies to x + 3 = 32, leading to x = 32 - 3 = 29. Thus, the correct answer is x = 29.

To solve 2log_6(x-8)=2, divide both sides by 2 to get log_6(x-8)=1. This means x-8=6^1, so x-8=6. Adding 8 gives x=14, which is the correct answer.

To solve log₆(x-3) = 2, we rewrite it in exponential form. This gives us 6² = x - 3, which is the correct choice. The other options do not correctly represent the logarithmic equation.