125^x + 5^(3x+1) = 400

Question

I've been stuck on this forever. I don't understand how to solve it with multiple terms with different bases. I have to solve for x

Answer 1

125^x + 5^3x+1 = 400

 

125^x + 5(5^3x) = 400

 

125^x + 5[(5³)^x] = 400

 

125^x + 5(125)^x = 400

 

125^x(1+5) = 400

 

6(125^x) = 400

 

125^x = 66.666666

 

log(125^x) = log(66.666666)

 

x(log(125)) = log(66.666666)

 

x = log(66.666666)/log(125) ≈ 0.87

 

Answer 2

125^x + 5^3x+1 = 400

(5³)^x + 5^3x+1 = 400

 

Can take it form here?

Answer 3

Rewrite the equation using common bases.

 

(5³)^x + 5^{(3x + 1)} = 400

 

5^3x + 5^3x(5) = 400

 

 

Subtract 400 on both sides of the equation.

 

5^3x + 5^3x(5) - 400 = 0

 

 

Let       r = 5^3x

 

Then,

 

r + 5r - 400 = 0

 

 

Solve the equation for r.  Then once you found your value of r, substitute back into the original substitution.

 

r = 5^3x

 

 

Then you will log both sides of the equation and bring down any exponents as the coefficient of the log.

 

Log(r) = 3xLog(5)

 

Log(r) / [3Log(5)] = x     -------> solution