Let logac and Logbc be two of the equations x ^ 2-3x + 1 = 0, and find the value of loga / BC

Let logac and Logbc be two of the equations x ^ 2-3x + 1 = 0, and find the value of loga / BC

Log (A / b) [C] = 1 / {log (c) [A / b]} and: log (c) [A / b] = log (c) [a] - log (c) [b] = 1 / {log (a) [C]} - 1 / {log (b) [C]}. = (1 / x1) - (1 / x2), then: log (A / b) [C] = 1 / [(1 / x1) - (1 / x2)] = (x1x2) / (x2-x1) can be calculated and substituted

Let loga (c) and logb (c) be two of the equations x ^ 2-3x + 1 = 0, and find the value of logab (ABC)
Do not use the value of loga / b (c) to find logab (ABC)

Log (AB) ABC = log (AB) AB + log (AB) C = 1 + 1 / log (c) AB = 1 + 1 / [log (c) a + log (c) b] = 1 + 1 / [1 / log (a) C + 1 / log (b) C] = 1 + log (a) C × log (b) C / [log (a) C + log (b) C] by Weida's theorem log (a) C + log (b) C = 3, log (a) C × log (b) C = 1 with the original formula = 1 + 1 / 3 = 4 / 3

Let loga (m) and logb (m) be two of the equations x ^ 2-3x + 1 = 0, and try to find the value of loga (m)
The answer is + radical 5 / 5 or - radical 5 / 5

In this paper, the logm (a (m) + logb (m) = 3 = 1 / logm (a) + 1 / logm (b), loga (m) * logb (m) = 1 / [logm (a) * logm (b)] and the first expression is the first expression to get logm (a) + logm (b (m) (logm (a) + logb (m (m) (logm (a) + logb (m (m) (logm (a) + logb (logm (logm (a) * logm (logm (a) (logm (a) (logm (a) (logm (a) (logm (a) (A / b) logm (A / b) | logm (A / b) logm (a) (logm (a) - logm (a) - logm (a) - logm (a) - logm (a) - logm (a) - logm (a) - logm (a) - logb) = 1 / radical (5) or - 1 / radical (5)