If log1/√2(1/√8)=log2(4x+1).log2(4x+1+4), then x=.................
Given: log1/√2(1/√8)=log2(4x+1).log2(4x+1+4)
⇒log1/√2(1/√2)3=log2(4x+1).log2(4x+1+4)
⇒3=log2(4x+1).log2(4x+1+4)
Since product of two no is 3, so one number will be 1 and other number will be 3.
⇒1×3=log2(4x+1).log2(4x+1+4)
Case 1: log2(4x+1)=1
⇒4x+1=2⇒4x=1⇒x=0
Now for x=0; second number should be 3.
So,
log2(4x+1+4)=log2(40+1+4)=log28=3
Hence, x=0 is satisfy the above equation.