Wipro WILP Logarithms Quiz- 1

log ₇ x = 9 ⇒ x = 7⁹.
∴ log _x y = 343
⇒ y = x³⁴³
⇒ y = (7⁹)³⁴³ [put value of x]
⇒ y = 7³⁰⁸⁷.

log(pq⁵) - log(pr) + log(pqr⁹) - 7log(qr)
= log(pq⁵) - log(pr) + log(pqr⁹) - log(q⁷r⁷)
= log_p(pq⁵/pr)*(pqr⁹/q⁷r⁷)
= log_p qr

Given Exp. = log [{(25/4) / (125/16)} x 5/4)]
= log [(25/4) x (5/4) x (125/16)]
= log 1
= 0

Given Exp. = log72/9 - log96/125 + log36/250
= log [(24 x 3) / (3 x 3)] - log (24/125) + log [(9 x 4) / (125 x 2)]
= log(24 x 3) - log ( 9 ) - log(24) + log125 + log(9 x 4) -log (125 x2)
= log 24 + log 3 - log 9 - log 24 - log 4 + log 125 + log 9 + log 4 - log 125 - log 2
= log 3 - log 2

= log (3/2)

logX+logY=log(XY)
log(x⁴/y²z²)+log(y⁴/x²z²)+log(z⁴/x²y²)
=log(x⁴ y⁴ z⁴ /z²x²z²x²y²)
=log(x⁴ y⁴z⁴/ x⁴y⁴z⁴)
=log1=0

Since we given that

log_b (p) + log_b (1+p) = 0

∵ Log(a×b) = Log(a) + Log(b)

∴ log_b ((p)(1+p))=0

∵ According to the definition of logarithms at the base of b

Logb(a) = c ⇔ b^c = a

So

b^(0) = ((p)(1+p))

1 = p² + p ∵ b^(0) = 1

By rearranging the terms we get

p² + p - 1 = 0

Thus

p² + p - 1 = 0 is equivalent to log_b (p) + log_b (1+p) = 0

Ans is p² + p - 1 = 0

1 + 2 + 3 + 4 + 5
= 15

log₇(2401)=xlog₇(2401)=x
Rewrite the equation as x=log₇(2401)x=log₇(2401).
x=log₇(2401)x=log₇(2401)
Logarithm base 7 of 2401 is 4,.
x=4

If log 14641 / log 11 = x
then x = 4log11/ log11= 4