5^2x+20*5^x-125=0
5^2x = (5^x)^2
So, make things easier to look at by letting
u = 5^x
Then you have
u^2+20u-125 = 0
(u+25)(u-5) = 0
u = -25 or 5
But, 5^x is never negative, so we end up with
u=5
5^x = 5
x = 1
check:
5^2 + 20*5 -125 = 25+100-125 = 0
5^a + 20*5^b = 0
a log 5 + log 20 + b log 5
To solve the equation 5^(2x) + 20 * 5^x - 125 = 0, we can use a substitution to simplify the equation.
Let's introduce a new variable: u = 5^x.
Substituting u into the equation, we get: u^2 + 20u - 125 = 0.
Now, we have a quadratic equation in terms of u. We can solve this quadratic equation using various methods, such as factoring, completing the square, or using the quadratic formula.
In this case, let's use factoring:
(u + 25)(u - 5) = 0.
Now we have two possible solutions:
1) u + 25 = 0 --> u = -25
2) u - 5 = 0 --> u = 5
Recall that we introduced the substitution u = 5^x. Substitute u back into the equation to find the values of x.
For u = -25:
5^x = -25.
However, 5^x represents an exponential function, and exponential functions are always positive. So, there are no solutions for this case.
For u = 5:
5^x = 5.
Using logarithms, we can solve for x:
Take the logarithm (base 5) of both sides:
log5(5^x) = log5(5).
Applying the logarithmic property logb(b^x) = x:
x = 1.
Therefore, the equation 5^(2x) + 20 * 5^x - 125 = 0 has one solution: x = 1.