How do you solve #5^(x^2+2x)=125#?

2 Answers
Dec 2, 2016

Answer:

#x=-3,x=1#

Explanation:

To solve the equation #5^(x^2+2x)=125# for variable #x#, we must apply the logarithm conversion formula.

logarithm to exponential conversion: #\log_a(b)=x\leftrightarrow a^x=b#


calculations

  • #5^(x^2+2x)=125\rArr\log_5(125)=x^2+2x#
  • You're going to need a calculator to solve the left side's logarithm, but you should get the following:
    #3=x^2+2x#
  • Now subtract 3 from both sides to form a quadratic equation:
    #0=x^2+2x-3#
  • You should get roots of -3 and 1. #\rArr(x+3)(x-1)#

Therefore, #x=-3,1#

Dec 4, 2016

Answer:

#x=-3# or #x=1#

Explanation:

Another way to approach this is to realize that #125=5^3#. Then we see that

#5^(x^2+2x)=5^3#

We now have two equal bases, each to a power. Since these are equal, we can say that their exponents must be equal. (We could write a rule for this and say that if #a^b=a^c#, then #b=c#).

So we know that

#x^2+2x=3#

Solving like a regular quadratic:

#x^2+2x-3=0#

#(x+3)(x-1)=0#

So #x=-3# or #x=1#.