Evaluate and solve: log 5 + log 16 - 3log 2, 2log 3 - 2log 2, log10(2x-5)=1, log(x+2)+log(x-2)=log 60

You're on a roll — let's evaluate these logarithmic expressions and solve for x. Evaluate: 1) 5 + 16 - 3 2 Step 1: Apply the power rule of logarithms, n a = a^n. 5 + 16 - 2^3 5 + 16 - 8 Step 2: Apply the product rule of logarithms, a + b = (ab). (5 × 16) - 8 80 - 8 Step 3: Apply the quotient rule of logarithms, a - b = ((a)/(b)). ((80)/(8)) 10 Step 4: Since the base of is usually 10 when not specified, _10 10 = 1. 1 The value is 1. 2) 2 3 - 2 2 Step 1: Apply the power rule of logarithms, n a = a^n. 3^2 - 2^2 9 - 4 Step 2: Apply the quotient rule of logarithms, a - b = ((a)/(b)). ((9)/(4)) The value is ((9)/(4)). Find the value of x: 1) _10(2x - 5) = 1 Step 1: Convert the logarithmic equation to an exponential equation. If _b a = c, then b^c = a. 10^1 = 2x - 5 10 = 2x - 5 Step 2: Solve the linear equation for x. 10 + 5 = 2x 15 = 2x x = (15)/(2) The value of x is (15)/(2). 2) (x+2) + (x-2) = 60 Step 1: Apply the product rule of logarithms, a + b = (ab), to the left side. ((x+2)(x-2)) = 60 Step 2: If A = B, then A=B. Equate the arguments of the logarithms. (x+2)(x-2) = 60 Step 3: Expand the left side using the difference of squares formula (a+b)(a-b) = a^2 - b^2, and solve the quadratic equation. x^2 - 4 = 60 x^2 = 60 + 4 x^2 = 64 x = ± sqrt(64) x = ± 8 Step 4: Check for valid solutions. The argument of a logarithm must be positive. For x=8: (8+2) = 10 > 0 and (8-2) = 6 > 0. This is a valid solution. For x=-8: (-8+2) = -6 < 0. This is not a valid solution. The value of x is 8. 3) ( 32)/( 128) = x Step 1: Express 32 and 128 as powers of the same base, which is 2. 32 = 2^5 128 = 2^7 Substitute these into the equation. x = ( 2^5)/( 2^7) Step 2: Apply the power rule of logarithms, a^n = n a. x = (5 2)/(7 2) Step 3: Cancel out 2 from the numerator and denominator. x = (5)/(7) The value of x is (5)/(7). What's next?