# solve log4 x + logx 4 = 2?

• solve log4 x + logx 4 = 2?  Answer #1 | 31/12 2013 02:04 http://www.coolmath.com/algebra/17-exponentials-logarithms/index.html
Positive: 75 %
Answer #2 | 31/12 2013 03:25 Given log_4 x + log_x 4 = 2 Change the base =>log_4(x) + log_4(4)/log_4(x) = 2 Take LCM =>(log_4(x))^2+1=2 log_4(x) =>log_4(x)^2+1-2log_4(x)=0 This is in the form of a^2+b^2-2ab=(a-b)^2 =>(log_4(x)-1)^2=0 =>log_4(x)-1=0 =>log_4(x)=1 =>x=4^1 =>x=4
Positive: 66.666666666667 %
Answer #3 | 31/12 2013 03:07 log_4 x + log_x 4 = 2. Use the change of base formula and properties of logs to get all terms in terms of base 4 logs. log_4 x + log_4 4/log_4 x = 2. Now multiply through by log_4 x (log_4 x)² - 2log_4 x + 1 = 0. Now factor (log_4 x - 1)² = 0 log_4 x = 1 x = 4
Positive: 60 %
Answer #4 | 31/12 2013 05:34 x=4 see step by step solution: http://symbolab.com/solver/equation-calculator/log_%7B4%7D%20x%20%2B%20log_%7Bx%20%7D4%20%3D%202 hope this helps
Positive: 60 %
Answer #5 | 31/12 2013 05:06 log x + log[x] 4 = 2 => Consider this rule in logarithm: log[a] b = 1/log[b] a, then by its application, you'll have log x + log[x] 4 = 2 → log x + 1/log x = 2 → (log² x + 1)/log x = 2 → log² x + 1 = 2log x → log² x - 2log x + 1 = 0 => Let log x be p, then p² - 2p + 1 = 0 p² - p - p + 1 = 0 p(p - 1) - 1(p - 1) = 0 (p - 1)(p - 1) = 0 (p - 1)² = 0 p - 1 = 0 p = 1. => Remember that p is log x, then log x = 1 => Consider this rule in logarithms: log[a] b = c in index form is b = a^c, then by its application, you'll have log x = 1 → 4 = x¹ Therefore, x = 4 ...Ans.
Positive: 57.142857142857 %
Answer #6 | 31/12 2013 02:37 Log4 x. + Logx 4 = 2 log 4 / log x + log x / log 4 = 2 (log 4)^2 + (log x)^2 - 2 log 4 log x = 0 let log x = y and log 4 = a y^2 - 2ay + a^2 = 0 y = a => log x = log 4 => x = 4
Positive: 50 %
Answer #7 | 31/12 2013 02:16 log_4(x) + log_x(4) = 2 log_4(x) + log_4(4)/log_4(x) = 2 ← change of base rule log_4(x)² + 1 = 2log_4(x) ← log_b(b) = 1 log_4(x)² - 2log_4(x) + 1 = 0 (log_4(x) - 1)² = 0 log_4(x) = 1 x = 4^1 x = 4
Positive: 50 % 