We are asked to solve `5^(2x)+20*5^x-125=0 ` :
Rewrite as ` (5^x)^2+20*5^x-125=0 ` and let `y=5^x ` to get
` y^2+20y-125=0` and
(y+25)(y-5)=0 so y=-25 or y=5.
y cannot be -25 as `5^x>0 ` for all real x.
If y=5 then x=1. If x=1 then 25+20(5)-125=0.
x=1 is the solution.
The graph:
We are asked to solve `5^(2x)+20*5^x-125=0 ` :
Rewrite as ` (5^x)^2+20*5^x-125=0 ` and let `y=5^x ` to get
` y^2+20y-125=0` and
(y+25)(y-5)=0 so y=-25 or y=5.
y cannot be -25 as `5^x>0 ` for all real x.
If y=5 then x=1. If x=1 then 25+20(5)-125=0.
x=1 is the solution.
The graph:
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