CoCubes Logarithms Question and Answers Quiz-1

log_x (0.1) = -1/3

=>(0.1) = x^(-1/3)

=>10^(-1) = x^(-1/3)

=>10 = x^(1/3)

=>x = 10^3

=>x = 1000

log₁₀ 5 + log₁₀ (5x + 1) = log₁₀ (x + 5) + 1

log₁₀ 5 + log₁₀ (5x + 1) = log₁₀ (x + 5) + log₁₀ 10

log₁₀ [5 (5x + 1)] = log₁₀ [10(x + 5)]

5(5x + 1) = 10(x + 5)

5x + 1 = 2x + 10

3x = 9

x = 3

Let log be log to base 8
log [ x/6 ] = 1/3
x/6 = 8^(1/3)
x/6 = 2
x = 12

xy = x + y

xy - y = x

y (x - 1) = x

y = x / (x - 1)

log10(1/70)=log10 1- log 10 70
= 0-log10 (10*7)
=-((log10 10 )+(log10 7))
=- (1+a)

Log[(a+b)/3] = 0.5[log(a) + log(b)]
=>log[(a+b)/3] = 0.5*log(ab)
=>log[(a+b)/3] = log[sqrt(ab)]
=>(a+b)/3 = sqrt(ab)
=>a + b = 3*sqrt(ab)
=>a² + 2ab + b² = 9ab [squaring both side]
=>a² + b² = 7ab

Logx= log 3 + 2 log 2- (3/4) log 16.

= log3+log4- log 16^3/4

=log3+log4-log2^12/4

=log3+log4-log2³

=log(3*4/8)

=log(3/2)

Logx=log3/2

x=3/2

log₁₀₀₀₀ X = -1/4

=>X = 10000^(-1/4)

=>X=1/10

3- (1/2)log₃ 9

=3 - (1/2)log₃ 3²

=3 - (2/2)log₃ 3

=3 - log₃ 3

=3 - 1

=2

a) Since log_a a = 1, so, log₁₀ 10 = 1.

(b) log (2 + 3) = log 5

log (2 x 3) = log 6

So, "log (2 + 3)" Not equal to "log (2 x 3)"

(c) Since log_a 1 = 0, so log₁₀ 1 = 0.

(d) log (1 + 2 + 3) = log 6 = log (1 x 2 x 3) = log 1 + log 2 + log 3.