Question

PLEASE HELP!!!
solve for the unknown: 3^x=9^x+5, 27^4x=9^x+1, 5^x=25^3x+5.

Answer 1

Explanation:

`3^x=9^(x+5)\\\\3^x=\left(3^2\right)^(x+5)\qquad|\text{use}\ (a^n)^m=a^(nm)\\\\3^x=3^(2(x+5))\iff x=2(x+5)\qquad|\text{use the distributive property}\\\\x=2x+10\quad|\text{subtract}\ 2x\ \text{from both sides}\\\\-x=10\qquad|\text{change the signs}\\\\\huge\boxed{x=-10}`

`27^(4x)=9^(x+1)\\\\\left(3^3\right)^(4x)=\left(3^2\right)^(x+1)\\\\3^((3)(4x))=3^(2(x+1))\iff(3)(4x)=2(x+1)\\\\12x=2x+2\qquad|\text{subtract}\ 2x\ \text{from both sides}\\\\10x=2\qquad|\text{divide both sides by 10}\\\\x=(2)/(10)\\\\\huge\boxed{x=0.2}`

`5^x=25^(3x+5)\\\\5^x=\left(5^2\right)^(3x+5)\\\\5^x=5^(2(3x+5))\iff x=2(3x+5)\\\\x=(2)(3x)+(2)(5)\\\\x=6x+10\qquad|\text{subtract}\ 6x\ \text{from both sides}\\\\-5x=10\qquad|\text{divide both sides by (-5)}\\\\\huge\boxed{x=-2}`