Question

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

Answer 1

`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`

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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`

Answer 2

`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`.