# Can someone please explain how to solve one or 2 of these exponential equations? Thanks!

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### 3 Answers

(3) `5^(x-4)=25^(x-6)`

To solve exponential equation, it is better if we can express each side with the same base.

`5^(x-4)=(5^2)^(x-6)`

Then, simplify the right side by applying the exponent rule `(a^m)^n = a^(m*n)` .

`5^(x-4)=5^(2x-12)`

Now that each side have the same base, set the exponents equal to each other.

`x-4=2x - 12`

And, isolate the x.

`x-2x=-12+4`

`-x=-8`

`x=8`

**Hence, the solution is x = 8.**

(4) `7^(3x+4)=49^(2x+1)`

Factor 49 to see if both if both sides can be expres with same base.

`7^(3x+4)=(7^2)^(2x+1)`

`7^(3x+4)=7^(4x+2)`

Since they can both be express with same base, to solve, set their exponents equal to each other.

`3x +4=4x + 2`

And, isolate x.

`3x-4x=2-4`

`-x=-2`

`x=2`

**Thus, the solution is x=2.**

6) `27^(4x-1) = 9^(3x+8)`

The trick for this problem is to recognize that 27 = 9^(3/2).

`(9^(3/2))^(4x-1) = 9^(3x+8)`

Using the exponent rule we can simplify

`9^(6x - 3/2) = 9^(3x+8)`

Now we set the exponents equal to each other because they have the same base.

`6x - 3/2 = 3x + 8`

`3x =19/2`

`x = 19/6`

7)

Here we need to recognize that 64 = 4^3

`4^(2x-5) = (4^3)^(3x)`

`4^(2x-5) = 4^(9x)`

`2x - 5 = 9x`

`-5 = 7x`

`-5/7 = x`

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