Logarithms:

There exist a positive number k such that $$\log_{2}x+\log_{4}x+\log_{8}x=\log_{k}x$$for all positive real numbers x.
If $$k=\sqrt[b]{a}$$ where a,b are natural nos., the smallest possible value of (a+b) is equal to-

Ans.75 (How????)

Question

Logarithms:

There exist a positive number k such that $$\log_{2}x+\log_{4}x+\log_{8}x=\log_{k}x$$for all positive real numbers x. 
If $$k=\sqrt[b]{a}$$ where a,b are natural nos., the smallest possible value of (a+b) is equal to-

Ans.75 (How????)

anonymous · Answer

@nitz

ganeshie8 · Answer

$\huge log_{b^n}a = \frac{log_ba}{n}$

ganeshie8 · Answer

equation simplifies to,

$\huge  \frac{log_2x^{11}}{6} = log_kx$

ganeshie8 · Answer

$\huge log_2 x  = \frac{6}{11}log_kx$

ganeshie8 · Answer

$\huge log_2 x = \frac{log_kx}{11/6}$

$\huge log_2 x = log_{k^{(11/6)}}x$

$\huge 2 = k^{11/6}$
$\huge k = 2^{6/11}$

b = 11, a = 64

ganeshie8 · Answer

smallest possible value for (a, b) is (2^6, 11) = (64, 11)

6/11 = 12/22, 
so, next immediate value for (a, b) is (2^12, 11*2) = (4096, 22)

anonymous · Answer

thanks

anonymous · Answer

its done already