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

$$\mathrm{solve}\mathrm{for}\mathrm{a}$$$$$$$$5\log {}_4\mathrm{a}+48\log {}_{\mathrm{a}}4=\frac{\mathrm{a}}{8}$$

Answered by 433 last updated on 14/Jul/17

$$$$$$$$$$5\log {}_{4}a+48\frac{\log {}_{4}4}{\log {}_{4}a}=\frac{a}{8}$$$$5\log {}_{4}a+\frac{48}{\log {}_{4}a}=\frac{a}{8}$$$$\log {}_{4}a=x\iff a=4^x$$$$5x+\frac{48}{x}=\frac{4^x}{8}$$$$5x+\frac{48}{x}=2^{2x-3}\left(1\right)$$$$x=4\Rightarrow 20+\frac{48}{4}=2^5$$$$If{x}_0<0wassolution$$$$5x_0+\frac{48}{x_0}<0\mathrm{\&}{2}^{2x_0-3}>0$$$$x>0$$$$f\left(x\right)=5x+\frac{48}{x}-2^{2x-3}$$$$f^{\prime }\left(x\right)=5-\frac{48}{x^2}-2^{2x-3}\times \ln 2\times 2$$$$0<x<3$$$$\frac{48}{x^2}+2^{2x-1}\times \ln 2>\frac{48}{x^2}+\frac{\ln 2}{2}>\frac{48}{9}+\frac{\ln 2}{2}>5$$$$x>3$$$$\frac{48}{x^2}+2^{2x-1}\times \ln 2>\frac{48}{x^2}+32\times \ln 2>5$$$$f^{\prime }\left(x\right)<0$$$$\left(1\right)has⩽1solution$$$$x=4\Rightarrow a=4^4$$

Answered by ajfour last updated on 14/Jul/17

$$\mathrm{a}=256$$$$\mathrm{let}\mathrm{a}=4^{\mathbf{x}},\mathrm{then}\log {}_4\mathrm{a}=\mathrm{x}$$$$\Rightarrow 5\mathrm{x}+\frac{48}{\mathrm{x}}=\frac{4^{\mathbf{x}-1}}{2}$$$$\Rightarrow \mathrm{x}=4(\mathrm{trial}\mathrm{and}\mathrm{error}\mathrm{with}$$$$\mathrm{derivative}\mathrm{test})$$$$\Rightarrow \mathrm{a}=4^4=256\left(\mathrm{only}\mathrm{one}\mathrm{solution}\right).$$$$$$

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