Logarithm Quiz-1

\log_{x}\frac{9}{16}=-\frac{1}{2}

or, x^{-\frac{1}{2}}=\frac{9}{16}

or, x=\frac{256}{81}

\log_{2}x=10

or, x=2^{10}

Now,

\log_{x}y=100

or,  y=x^{100}

or , y=2^{1000}

\log (2^{64})

= 64\log 2

= 19.26592

So, the characteristic is 19 .
Therefore, the number of digits in 2^{64} is 20.

\log_{10}\frac{1}{70}

= \log_{10}1-\log_{10}70

= -(\log_{10}7+\log_{10}10)

= -(1+a)

\log_{10}5+\log_{10}(5x+1)=\log_{10}(x+5)+1

or, \log_{10}5+\log_{10}(5x+1)=\log_{10}(x+5)+\log_{10}10

or, \log_{10}[5(5x+1)]=\log_{10}[10(x+5)]

or, x=3

\log_{9}59049

= \log_{9}9^{5}

= 5

In the given question , x, y and z are the sides of a right angled triangle,

Thus , z^{2}=x^{2}+y^{2}

Now,

\frac{1}{\log_{x+z}y}+\frac{1}{\log_{z-x}y}

= \log_{y}(x+z)+\log_{y}(z-x)

= \log_{y}(z^{2}-x^{2})

= \log_{y}(y^{2})

= 2

\log_{5}16,\log_{5}(3^{x}-4),\log_{5}(3^{x}+\frac{97}{16})are in arithmetic progression

Thus ,

2\log_{5}(3^{x}-4)=\log_{5}16+\log_{5}(3^{x}+\frac{97}{16})

or, (3^{x}-4)^{2}=16(3^{x}+\frac{97}{16})

Substituting  3^{x} as y and solving the above equation , we get,

y = 27

or, 3^{x}=27

or, x = 3

(\log_{2}x)^{2}-\log_{2}x^{4}-32=0

or, (\log_{2}x)^{2}-4\log_{2}x-32=0

Substituting \log_{2}x as y and solving the equation ,

we get,    y = 8, -4

Now , \log_{2}x = 8
or , x = 256

And, \log_{2}x = -4
or, x = 1/16

Since ‘x’ is an integer,  so ,
x = 256.

\log_{10}(2^{x}+x-41)=x(1-\log_{10}5)

or, \log_{10}(2^{x}+x-41)=x\log_{10}2

or, (2^{x}+x-41)=2^{x}

or, x = 41