Verified answer

Log5 (x^2 ) – Logx (5)

= 2 log5(x) - 1/log5(x) as loga(b) = 1/logb(a)

= 2y - 1/y = - 1 taking log5(x) = y

so 2y^2-1 = -y

or 2y^2+ y - 1 = 0

or (2y-1)(y+1) = 0

y = 1/2 or - 1

so x = 5^(1/2) or 5^-(1) i.e 1/5

Do you mean Log5 (x^2) - Logx (5)?

Log5 (x^2) - Logx (5)

=2Log5 (x) - Logx (5)

=2Log5 (x) - 1/ Log5 (x)

so Log5 (x2 ) – Logx (5) don't equals to -1 unless Log5 (x)=-1or 1/2

that means Log5 (x2 ) – Logx (5) don't equals to -1 unless x=sqrt(5) or x=1/5

Log5 (x2 ) – Logx (5) = -1

Log5 (x2 ) – Logx (5) = -1

log5 + log(x^2) - logx -log5 = -1

2.logx - logx = -1

logx = -1

By definition of log :

(10)^(-1) = x

1/10 = x

Thus :

x = 0.1

Answer

log5 (x²) - logx (5) = -1

2 log5 (x) - 1/log5 (x) = -1

2 [log5 (x)]² + log5 (x) - 1 = 0

[2 log5 (x) - 1][log5 (x) + 1] = 0

(i)

2 log5 (x) - 1 = 0

log5 (x) = 1/2

x = 5^(1/2)

(ii)

log5 (x) + 1 = 0

log5 (x) = -1

x = 5^(-1) = 1/5

2log x/log5 -log 5/log x=-1

log x/log 5 =y

2y-(1/y)=-1

2y^2-1=-y

2y^2+y-1=0

y=[-1+&-(1+8)^1/2]/4=-1 & 1/2

log x /log 5 =-1

log x = -log 5

x=1/5 one answer

log x /log5 =1/2

log x =1/2log 5

x=\/''5'''