Aricent Logarithms Quiz-1

a^{x}= b^{y}
Taking log both sides,
log a^{x} =log b^{y}
or, xloga=ylogb
or, loga/logb=y/x

F(a)+f(b)=f(a+b) is operation............
where f(x)=log x

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)
a^2 + 2ab + b^2 = 9 ab
a^2 + b^2 = 7ab

log(x) = (1/2) log(y) => y = x^2
log(x) = (1/5) log(z) => z = x^5
Then x^4*y^5*z = x^19

Log10000x=-1/4
4x=-1/4
-x=16
:x=16

Given, 3-frac{1}{2}log_39

=3-frac{1}{2}log_33^2

we know, log_ab^n=nlog_ab

=3-frac{1}{2} imes2log_33

we know, log_aa=1

=3-1 imes1

= 2

hence, answer is 2

They are in H.P.
a, b and c are in GP ( b^2=ac)
b^2=ac
logb^2= log a + log c
{all log are To base n }
loga , log b and logc are in AP
so, 1/loga , 1/logb and 1/logc are in HP
applying base changing law of log
loga to base n = 1/ {logn to base a}
==> 1/ loga = log(base a ) n
similarly , 1/log( base b ) n
and 1/log (base c) n
then
(loga n , logb n ,logc n) are in H.P.

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X^5 = 30
1^5 = 1, 2^5 = 32

The inference is that
x is a number between 1 and 2

Log2X

X triples ecery second

log2 × 3X
log 6X

The answer is 4 can written as 4 cube