What is $x$ $x$ in $4^{2 x + 5} = 5^{2 - x}$ $4^(2x+5) = 5^(2-x)$ ?

Question

1088 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-26T16:14:56

$4^{2 x + 5} = 5^{2 - x} = 4^{2 x} \cdot 4^{5} = 5^{2} \cdot 5^{- x}$ $4^(2x+5)=5^(2-x)=4^(2x)*4^5=5^2*5^-x$

take natural logarithm both side

$2 x ln (4) + 5 ln (4) = 2 ln (5) - x ln (5)$ $2xln(4)+5ln(4)=2ln(5)-xln(5)$

$ln (4) (2 x + 5) = ln (5) (2 - x)$ $ln(4)(2x+5)=ln(5)(2-x)$

$\frac{2 x + 5}{2 - x} = \frac{ln (5)}{ln (4)}$ $(2x+5)/(2-x)=ln(5)/ln(4)$

$\frac{2 x + 5}{2 x - 4} = - \frac{1}{2} \frac{ln (5)}{ln (4)}$ $(2x+5)/(2x-4)=-1/2ln(5)/ln(4)$

$\frac{2 x - 4 + 9}{2 x - 4} = - \frac{1}{2} \frac{ln (5)}{ln (4)}$ $(2x-4+9)/(2x-4)=-1/2ln(5)/ln(4)$

$\frac{9}{2 x - 4} = - \frac{1}{2} \frac{ln (5)}{ln (4)} - 1$ $9/(2x-4)=-1/2ln(5)/ln(4)-1$

$\frac{2 x - 4}{9} = \frac{1}{- \frac{1}{2} \frac{ln (5)}{ln (4)} - 1}$ $(2x-4)/9=1/(-1/2ln(5)/ln(4)-1)$

$\frac{2 x - 4}{9} = - \frac{2}{\frac{ln (5)}{ln (4)} + 2}$ $(2x-4)/9=-2/(ln(5)/ln(4)+2)$

$x - 2 = - \frac{9}{\frac{ln (5)}{ln (4)} + 2}$ $x-2=-9/(ln(5)/ln(4)+2)$

$x = - \frac{9}{\frac{ln (5)}{ln (4)} + 2} + 2$ $x = -9/(ln(5)/ln(4)+2)+2$

$x = - \frac{18 ln (2)}{ln (5) + 4 ln (2)} + 2$ $x = -(18ln(2))/(ln(5)+4ln(2))+2$

What is xxx in 4^(2x+5) = 5^(2-x)42x+5=52−x4^(2x+5) = 5^(2-x)?