Question
Find value of x for log1/2(x-1)+log1/2(x+1)-log1/√2(7-x)=1
Answers
Steve
log1/2(x-1)+log1/2(x+1)-log1/√2(7-x)=1
You seriously need to learn how to use parentheses. Also, the final result will depend on the base of the logarithms. I shall assume base 10.
Assuming you meant
log(1/(2(x-1)))+log(1/(2(x+1)))-log(1/(√2(7-x)))=1
log (1/(2(x-1)) * 1/(2(x+1)) * √2(7-x)) = 1
1/(2(x-1)) * 1/(2(x+1)) * √2(7-x) = 10
I'll let you solve the cubic in any way you like, but somehow I feel the question has been garbled.
I suggest you type in your equation at wolframalpha.com and see how it interprets your text. Then use enough parentheses to make it come out right.
You seriously need to learn how to use parentheses. Also, the final result will depend on the base of the logarithms. I shall assume base 10.
Assuming you meant
log(1/(2(x-1)))+log(1/(2(x+1)))-log(1/(√2(7-x)))=1
log (1/(2(x-1)) * 1/(2(x+1)) * √2(7-x)) = 1
1/(2(x-1)) * 1/(2(x+1)) * √2(7-x) = 10
I'll let you solve the cubic in any way you like, but somehow I feel the question has been garbled.
I suggest you type in your equation at wolframalpha.com and see how it interprets your text. Then use enough parentheses to make it come out right.