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Use the change of base formula to approximate the solution to log0.515 = 1 – 2x. Round to the nearest hundredth.
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Assuming the equation is log_{0.5}(15) = 1 − 2x, use change of base:
log_{0.5}(15) = ln(15)/ln(0.5) = −ln(15)/ln(2) = −log_2(15).
Compute log_2(15): log_2(15) = log_{10}(15)/log_{10}(2) ≈ 1.176091259 / 0.301029996 ≈ 3.906890596.
So 1 − 2x = −3.906890596 → 2x = 1 + 3.906890596 = 4.906890596 → x ≈ 2.453445298.
Rounded to the nearest hundredth: x ≈ 2.45.
log_{0.5}(15) = ln(15)/ln(0.5) = −ln(15)/ln(2) = −log_2(15).
Compute log_2(15): log_2(15) = log_{10}(15)/log_{10}(2) ≈ 1.176091259 / 0.301029996 ≈ 3.906890596.
So 1 − 2x = −3.906890596 → 2x = 1 + 3.906890596 = 4.906890596 → x ≈ 2.453445298.
Rounded to the nearest hundredth: x ≈ 2.45.
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