can someone help me solve: given logb3 = 0.6826 and logb7 = 1.2091, evaluate logb21. completely lost.
can someone help me solve: given logb3 = 0.6826 and logb7 = 1.2091, evaluate logb21. completely lost. You need to know this basic rule of logarithms: logb(X·Y) = logbX + logbY and the fact that 21 = 3·7, to get logb(3·7) = logb3 + logb7 then since you are given that logb3 = 0.6826 and logb7 = 1.2091 logb(3·7) = logb3 + logb7 = 0.6826 + 1.2091 = 1.8917 Edwin
Solve for x (log3x)2 + log3x2 + 1 =...(answered by Edwin McCravy)
Write in exponential form: log100.001 = -3(answered by AnlytcPhil)
log2(5x + 2) = 4(answered by Edwin McCravy)