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[4] 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[4] x + log[x] 4 = 2
→ log[4] x + 1/log[4] x = 2
→ (log²[4] x + 1)/log[4] x = 2
→ log²[4] x + 1 = 2log[4] x
→ log²[4] x - 2log[4] x + 1 = 0
=> Let log[4] 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[4] x, then
log[4] 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[4] 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

Answer #1| 31/12 2013 02:0475 %