solve-for-a-5log-4-a-48log-a-4-a-8-

Question Number 18034 by chux last updated on 14/Jul/17

solve for a

5 \log_{4} a + 48 \log_{a} 4 = \frac{a}{8}

Answered by 433 last updated on 14/Jul/17

5 \log_{4} a + 48 \frac{\log_{4} 4}{\log_{4} a} = \frac{a}{8}

5 \log_{4} a + \frac{48}{\log_{4} a} = \frac{a}{8}

\log_{4} a = x \Leftrightarrow a = 4^{x}

5 x + \frac{48}{x} = \frac{4^{x}}{8}

5 x + \frac{48}{x} = 2^{2 x - 3} (1)

x = 4 \Rightarrow 20 + \frac{48}{4} = 2^{5}

I f x_{0} < 0 w a s s o l u t i o n

5 x_{0} + \frac{48}{x_{0}} < 0 & 2^{2 x_{0} - 3} > 0

x > 0

f (x) = 5 x + \frac{48}{x} - 2^{2 x - 3}

f^{'} (x) = 5 - \frac{48}{x^{2}} - 2^{2 x - 3} \times \ln 2 \times 2

0 < x < 3

\frac{48}{x^{2}} + 2^{2 x - 1} \times \ln 2 > \frac{48}{x^{2}} + \frac{\ln 2}{2} > \frac{48}{9} + \frac{\ln 2}{2} > 5

x > 3

\frac{48}{x^{2}} + 2^{2 x - 1} \times \ln 2 > \frac{48}{x^{2}} + 32 \times \ln 2 > 5

f^{'} (x) < 0

(1) h a s ⩽ 1 s o l u t i o n

x = 4 \Rightarrow a = 4^{4}

Answered by ajfour last updated on 14/Jul/17

a = 256

let a = 4^{\boldsymbol x}, then \log_{4} a = x

\Rightarrow 5 x + \frac{48}{x} = \frac{4^{\boldsymbol x - 1}}{2}

\Rightarrow x = 4 (trial and error with

derivative test)

\Rightarrow a = 4^{4} = 256 (only one solution) .

Facebook Tweet Pin