SOLUTION: solve for 'x' : 4logx-logx^3 +log2 = log(6-x)

Question 368023: solve for 'x' :
4logx-logx^3 +log2 = log(6-x)
Answer by robertb(5830) (Show Source): You can put this solution on YOUR website!
log(x^4) - log(x^3)+log2 = log(6-x).
log(x^4/x^3)+log2 = log(6-x),
logx+log2 = log(6-x),
log(2x) = log(6-x),
2x = 6-x, since log is a 1-to-1 function. Hence
3x = 6,
x = 2. This satisfies the original equation, and thus is the final answer.
RELATED QUESTIONS
Solve for "x" if... (answered by ikleyn)

What is logx+log(x-1)=log2? Solve for... (answered by stanbon)

log(x+6)+logx=3 solve for... (answered by josmiceli)

logx+log(x+1)=log2 (answered by josgarithmetic,ikleyn)

solve.... (answered by lwsshak3)

{{{ log3=4logx }}}. solve for... (answered by HyperBrain)

Express the following in index form Logx + logy =1 Logx =6 Log (x+4) = log64 Log... (answered by stanbon)

log (x^2+1)-logx =... (answered by ikleyn)

HELP =) ! resolving the Question 2570 (p200 - 250) : log of 9 to the base square root... (answered by rapaljer,venugopalramana)