Quiz-1

log ₇ x = 9 ⇒ x = 7⁹.
∴ log _x y = 343
⇒ y = x³⁴³
⇒ y = 7^{(9 x 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 x 16/4 x 125)]+ log5/4
= log [(4/5)x(5/4)]
= log 1 = 0

Given Exp. = log(72/9) - log(96/125) + log(36/250)
= log [(72 x 125)/(9 x 96)] + log (36/250)
= log(125/12) + log (36/250)
= log [(125 x 36)/(12 x 250)]
= log (3/2)

logx+logy=log(xy)
log(\frac{x^{4}}{y^{2}z^{2}})+log(\frac{y^{4}}{x^{2}z^{2}})+log(\frac{z^{4}}{x^{2}y^{2}})
=log[(x⁴ y⁴ z⁴⁾ /(z²x²z²x²y²)]
=log(x⁴ y⁴z⁴/ x⁴y⁴z⁴)
=log1=0(zero)

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

log2 (2) + log2 (2^2) + log2 (2^3) + log2 (2^4) + log2 (2^5)
As we know, logp (x^y) = y logp (x)
log2 (2) +2 log2 (2) +3 log2 (2) +4 log2 (2) +5 log2 (2)
15 log2 (2) =15

log₇ (1/2401)

=log₇ (1) - log₇ (2401)

=0 - log₇ (7^4)

=-4log₇ (7)

=-4

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

Explanation:
log(x)=\frac{1}{2}log(y)
y=x^{2}
log(x)=\frac{1}{5}log(z)
z=x^{5}

Then,x^{4}\times y^{3}\times z^{-2}
=x^{4}\times x^{6}\times x^{-10}[substituting the value of x from above]
=x^{10}\times x^{-10}
=x^{0}
=1