Advertisements
Advertisements
प्रश्न
Find x, if : logx (5x - 6) = 2
Advertisements
उत्तर
logx (5x - 6) = 2
⇒ 5x - 6 = x2 ...[ Removing logarithm ]
⇒ x2 - 5x + 6 = 0
⇒ x2 - 3x - 2x + 6 = 0
⇒ x( x - 3 ) - 2( x - 3 ) = 0
⇒ ( x - 2 )( x - 3 ) = 0
∴ x = 2, 3.
APPEARS IN
संबंधित प्रश्न
If `3/2 log a + 2/3` log b - 1 = 0, find the value of a9.b4 .
If x = 1 + log 2 - log 5, y = 2 log3 and z = log a - log 5; find the value of a if x + y = 2z.
Evaluate: logb a × logc b × loga c.
Solve : log5( x + 1 ) - 1 = 1 + log5( x - 1 ).
If a2 = log x , b3 = log y and `a^2/2 - b^3/3` = log c , find c in terms of x and y.
State, true of false:
logba =-logab
If 2 log x + 1 = log 360, find: log (3 x2 - 8)
If x + log 4 + 2 log 5 + 3 log 3 + 2 log 2 = log 108, find the value of x.
If a = `"log" 3/5, "b" = "log" 5/4 and "c" = 2 "log" sqrt(3/4`, prove that 5a+b-c = 1
Express the following in a form free from logarithm:
2 log x + 3 log y = log a
