How would you find 'x' in 5 x 5^(4X-2) = 25^(x+3)?

Question

in parenthesis, is the whole exponent. i have no clue how to find x!!

Answer 1

Whenever you see your variable in an exponent, you should automatically think of taking a log. Logs are how you get variables out of exponents.

But first, let's simplify a bit.

Since 25 is 5 squared, we can rewrite the equation as:

5 * 5^(4x-2) = (5^2)^(x+3)
or, applying a rule of exponents,
5 * 5^(4x-2) = 5^(2*(x+3))
which turns into
5 * 5^(4x-2) = 5^(2x+6)

Divide both sides by 5^(4x-2) to get
5 = (5^(2x+6)) / (5^(4x-2))
Which turns into
5 = 5^((2x+6)-(4x-2))
or
5 = 5^(-2x+8)

Now you're ready to take a log of both sides. You can use log, or natural log ln, or log base five, doesn't matter. I prefer ln, but they all give the same answer. (Logs are really neat that way!)

ln 5 = ln 5^(-2x+8)
And by one of the log rules, this turns into
ln 5 = (-2x+8) * ln 5
Divide both sides by ln 5 to get
1 = -2x + 8
And with some simple algebra, you get x = 3.5, which is the solution to your problem.

Answer 2

OK, first add 1 to the left side since 5*5 is 5^(1+1) this leaves you with 5^(4x-1) = 25^(x+3) then you figure that 25 is just 5^2 so you can make the equation 5^(4x-1) = 5^2(x+3) then since the 5^power is on both sides the equation is now (4x-1) = 2x+6 which leaves x=7/2.

Answer 3

The key here is to manipulate the powers of the the exponents.

(1) 5*5^(4x-2) = 25^(x+3)
(2) 5^(4x-2+1) = 5^(2*(x+3))
(3) 5^(4x-1) = 5^(2x+6)
(4) 4x-1 = 2x+6
(5) 2x = 7
Therefore, x = 7/2

Answer 4

Oh, wow. Exponents. Get rid of them. Take the base-5 log of both sides and solve that. Fortunately, the equation does not add or subtract except in exponents.

Answer 5

X is right there dude. you don't see it? theres like 3 of them.