Question

log2 base 10, log2^x-1 base 10 and log2^x+3 base 10 is in ap find x​

Answer 1

Question :

$\sf \log_{10}(2) , \: log_{10}(2 {}^{x - 1} ) \:, \ log_{10}(2 {}^{x + 3} )$ are in Ap then find the value of x .

Theory :

•If a,b,c are in AP

then 2b=a+c

•Properties of Logarithm:

$\sf1) \log(x) + \log(y) = log(xy)$

$\sf2) \ log(x {}^{n} ) = n \log x$

$\sf3) \log( \sf \frac{x}{y} ) = \log(x) - \log(y)$

Solution :

$\bf \log_{10}(2) \:, \ log_{10}(2 {}^{x - 1} ) \: ,\ log_{10}(2 {}^{x + 3} )$are in Ap

Therefore,

$\sf \implies2 \log_{10}(2 {}^{x - 1} ) = \log_{10}(2) + \log_{10}(2 {}^{x + 3} )$

We Know that :

$\sf \log x {}^{n} = n \log \: x$

$\sf \implies \log_{10}(2 {}^{2( x- 1)} ) = \log_{10}(2) + \log_{10}(2 {}^{x + 3} )$

$\sf \implies \log_{10}(2 {}^{2x - 2} ) = \log_{10}(2) + \log_{10}(2 {}^{x + 3} )$

We know that

$\sf \log(x) + \log(y) = \log(xy)$

$\sf \implies log_{10}(2 {}^{2x - 2} ) = log_{10}(2 \times 2 {}^{x + 3} )$

$\sf \implies log_{10}(2 {}^{2x - 2} ) = log_{10}(2 {}^{x + 4} )$

On camparing both sides

$\sf \implies2x - 2 = x + 4$

$\bf \implies \: x = 6$

Hence the value of x is 6.

_____________________

Verification:

$\sf log_{10}(2) \: , log_{10}(2 {}^{x - 1} ) \: , log_{10}(2 {}^{x + 3} )$

put x = 6

Thus,

$\sf log_{10}(2) \:, log_{10}(2 {}^{5} ) \: , log_{10}(2 {}^{9} )$

If they are in Ap

$\sf \implies 2 log_{10}(2 {}^{5} ) = log_{10}(2) + log_{10}(2 {9}^{2} )$

$\sf \implies10 log_{10}(2) = 10 log_{10}(2)$

Hence Verified, they are in Ap, if x = 6 .