Question

If log 8 a + log 4 b^2 =5 and log 8 b + log4 a^2=7 then find value of ab

Answer 1

$\sf{GIVEN:-}$ $\sf{\log_8a + \log_4b^2 = 5}$

$\sf{\frac{\log_2a}{\log_28} + \frac{\log_2b^2}{\log_24} = 5}$

$\sf{\frac{\log_2a}{\log_22^3} + \frac{\log_2b^2}{\log_22^2} = 5}$

$\sf{\frac{\log_2a}{3\log_22} + \frac{2\log_2b}{2\log_22} = 5}$

$\sf{\frac{\log_2a}{3} + \frac{2\log_2b}{2} = 5}$

$\sf{\frac{\log_2a}{3} + \log_2b= 5}$------------------( 1 )

$\Large{\bold{AND}}$ $\sf{\log_8b + \log_4a^2 = 7}$

$\sf{\frac{\log_2b}{\log_28} + \frac{\log_2a^2}{\log_24} = 7}$

$\sf{\frac{\log_2b}{\log_22^3} + \frac{\log_2a^2}{\log_22^2} = 7}$

$\sf{\frac{\log_2b}{3\log_22} + \frac{2\log_2a}{2\log_22} = 7}$

$\sf{\frac{\log_2b}{3} + \frac{2\log_2a}{2} = 7}$

$\sf{\frac{\log_2b}{3} + \log_2a = 7}$-------------( 2 )

$\sf{[put \log_2b = x , \log_2a = y]}$

now the equation will be ,

y/3 + x = 5

(y + 3x)/3 = 5

3x + y = 15--------------( 1 )

AND,

x/3 + y = 7

(x + 3y)/3 = 7

x + 3y = 21----------( 2 )

from----( 1 )
y = (15 - 3x) [ put in ----( 2 )]

x + 3(15 - 3x) = 21

x + 45 - 9x = 21

-8x = 21 - 45

-8x = -24

x = -24/-8 = 3 [ put in -----( 1 )]

3x + y = 15

3(3) + y = 15

y = 15 - 9

y = 6 , x = 3

now, $\sf{x = \log_2b = 3}\\ \\ \sf{b = 2^3 = 8}$

$\sf{y = \log_2a = 6}\\ \\ \sf{b = 2^6 = 64}$

$\huge{\sf{HENCE, a*b = 8 * 64 }}\\ \\ \huge{\sf{a*b = 512}}$

Answer 2

Answer:

Step-by-step explanation:

answer is 2 to the power 9 = 512