Question

Let a, b and c be real numbers, each greater than 1, such that 2/3 log of a to the base b+3/5 logb to the base c+ 5/2 logc to the base a=3. If the value of b is 9, then the value of a must be what?

Answer 1

Answer with explanation:

It is given that, a, b and c be real numbers, each greater than 1.

The given equality consisting log is

$\frac{2}{3}\log_{b}a+\frac{3}{5}\log_{c}b+\frac{5}{2}\log_{a}c=3\\\\ b=9\\\\\log_{9}a^{\frac{2}{3}}+\log_{c}9^{\frac{3}{5}}+\log_{a}c^{\frac{5}{2}}=3$

As, there are two unknown variables , a and c,considering the above equation as Identity

$\log_{9}a^{\frac{2}{3}}=1\\\\\frac{\log a^{\frac{2}{3}}}{\log 9}=1\\\\ \log a^{\frac{2}{3}}=\log 9\\\\ a^{\frac{2}{3}}=9\\\\ a=9^{\frac{3}{2}}\\\\ a=(3^2)^{\frac{3}{2}}\\\\ a=3^3\\\\a=27$

Similarly,

$\log_{c}9^{\frac{3}{5}}=1\\\\ \log 9^{\frac{3}{5}}=\log c\\\\ \log (3^2)^{\frac{3}{5}}=\log c\\\\ c=3^{\frac{6}{5}}$

Answer 2

Answer:

27

Step-by-step explanation:

Let a, b and c be real numbers, each greater than 1, such that 2/3 log of a to the base b+3/5 logb to the base c+ 5/2 logc to the base a=3. If the value of b is 9, then the value of a must be what?

$\frac{2}{3} \log_{b}a + \frac{3}{5} \log_{c}b + \frac{5}{2}\log_{a}c = 3$

Now

$\frac{5}{2} \log_{a}c = \frac{5}{3} \times \frac{3}{2} \times\frac{\log_{b}c}{\log_{b}a}$

$\log_{b}c = \log_{c}b\\ \\ \frac{5}{2} \log_{a}c = \frac{5}{3} \times\frac{3}{2} \times\frac{1}{\log_{b}a\log_{b}c}\\ \\ \frac{5}{2}\log_{a}c =\frac{1}{\frac{3}{5}\log_{b}a \times \frac{2}{3}\log_{b}c}$

$\frac{2}{3} \log_{b}a =x \\ \frac{3}{5} \log_{c}b = y$

then

x + y + 1/xy = 3

x & y are positive number as a , b , c > 1

This is only true for positive number of x = y = 1

$\frac{2}{3} \log_{b}a =1$

$\log_{b}a = \frac{3}{2}$

$b^{\frac{3}{2}} = a$

b = 9

$9^{\frac{3}{2}} = a$

$(3^2)^{\frac{3}{2}} = a$

$3^3 = a$

a = 27