Given `e^(x^2+6)=e^(5x)` . We have to solve for x.

The one-to-one property is given by,

If `a^m=a^n` then m=n.

So here,

`e^(x^2+6)=e^(5x)`

Implies, `x^2+6=5x`

i.e `x^2-5x+6=0`

` x=2,3`

Since the entire base on the left side of the equation is equal to the entire base on the right side of the equation (in other words e = e), then the powers on the left must equal the powers on the right. So, x^2 + 6 = 5x. Move the 5x from the right side to the left side (change sides, change signs) and then you will have:

x^2 - 5x + 6 = 0. From there you must factor, getting (x - 3)(x - 2) = 0. And from there it's a simple step to get x = 3 and x = 2.

# Use the One-to-One Property to solve the equation for x.

The one-to-one property allows us to set powers equal to each other when we have the same base. Both bases on the left and right sides of the equation are " `e`".

Set the powers equal to each other.

`x^2 +6 = 5x`

Subtract `5x` from both sides.

`x^2-5x+6 = 0`

Factorize the left side.

`(x-3)(x-2)=0`

The roots are:

`x= 2,3`

You can plug both solutions back to the original equation, which satisfy both sides of the equation.

` `

We have `e^f(x)=e^g(x) ` which means that `f(x)=g(x) ` , where `f(x)=x^2+6 ` and `g(x)=5x `.

To find the value of `x ` such that `x^2+6=5x ` we may plot the difference `x^2-5x+6=0 `:

It is easy to verify that the solution to this quadratic equation is `x={2,3} `. Of course, you could have used the quadratic formula, or factored the equation into `(x-2)(x-3)=0 `.

So we are given

Since both of these problems are e raised to some power, we can just compare the exponents, and set them equal to each other.

So now we have `x^2+6=5x`

If we subtract 5x from both sides then we will be left with:

`x^2-5x+6=0 `

Then we can factor the left side of this equation:

`(x-2)(x-3)=0`

Now we can split this up to two separate equalities:

`x-2=0 `

AND

` x-3=0`

We can do this because then we would either have:

`(0)(x-3)=0` or `(x-2)(0)=0`

So then we would solve both equalities for x, and would have:

`x=2` or `x=3`

We can solve the problem this way because then we would have one of these equalities when we plug in these values of x that we found:

`(0)(-1)=0` or `(1)(0)=0`

The equation must be solved using one to one property of bijective functions, hence, since the base is the same both sides, then the powers must be set equal, such that:

`x^2 + 6 = 5x`

You need to subtract 5x both sides, such that:

`x^2 - 5x + 6 = 0`

You need to use quadratic formula:

`x_(1,2) = (-b +- sqrt(b^2 - 4*a*c))/(2*a)`

`a = 1, b = -5, c = 6`

`x_(1,2) = (5 +- sqrt(5^2 - 4*6*1))/(2*1)`

`x_(1,2) = (5 +- sqrt(25 - 24))/2`

`x_(1,2) = (5 +- 1)/2`

`x_1 = 3 ; x_2 = 2`

**Hence, the solutions to the exponential equation are `x_1 = 3 ; x_2 = 2` .**

Taking natural log of both sides,

`ln(e)^(x^2+6) = ln(e)^(5x)`

or, `(x^2+6) ln(e)= 5x ln(e)`

or, `x^2 +6 = 5x`

or, `x^2 -5x +6 = 0`

`x^2 -3x -2x +6 = 0`

`(x-2)(x-3) = 0`

`x= 2 or x = 3`

The given equation is :-

e^{(x^2) + 6} = e^(5x)

Taking log to the base 'e' both sides we get,

(x^2) + 6 = (5x)

or, (x^2) - 5x + 6 = 0

or, {(x^2) - 2x} + {6 - 3x} = 0

or, x(x-2) - 3(x-2) = 0

or, (x-2)*(x-3) = 0

Thus, **either x = 2 or, x = 3**