I'm really bad at logorithms. given the exponential equation 2x = 128, what is the logarithmic form of the equation in base 10?

Sorry to hear that you already took the test. Here is how to do the problem:

2ˣ = 128               
Let's take log base 10 of both sides.
For simplicity, we'll just use the symbol log to mean log₁₀

log (2ˣ) = log (128)                   Remember that log(yˣ) = x log(y)
x log (2) = log (128)                 Divide both sides by log(2) to get x by itself
x = log (128) ÷ log (2)

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12