(125/27)^x+1=(3/5)^x-1 ???

Question

The answer is (-1/2) , I just need to know how to get the answer please! Thank you!

Answer 1

(125/27)^x+1=(3/5)^x-1

You can apply ln in both sides..

ln(125/27)^x+1 = ln(3/5)^x-1

(x+1) ln 125/27 = (x-1) ln 3/5

After solving logartims

(x+1) / (x-1) = -1/3

3x+3 = -x+1

4x= -2

x= -1/2

Answer 2

Here's the (not so) quick and (quite) dirty way:
I'm going to simplify the equation first to save on typing:

A^x+1 = B^x-1 (A=125/27, B=3/5)

This is the same as:

(A^x-1)(A^2) = B^x-1

Divide both sides by A^x-1

A^2 = B^x-1 / A^x-1
A^2 = (B/A)^x-1

What just happened there: We don't know what x is, but we know that the formula calls for B/A to happen exactly x-1 times. Therefore, we can simplify that way.

Now simplify the B/A a little:

A^2 = ((B)(1/A))^x-1

Now I'll stick A and B back in:

(125/27)^2 = ((3/5)(27/125))^x-1
(5^3/3^3)^2 = (3^4/5^4)^x-1
5^6/3^6 = (3^4/5^4)^x-1
0 = 3^6/5^6 (3^4/5^4)^x-1

For this to be true, (x-1) has to change the 4s to _negative_ 6s. When that happens, you'll have the formula resolve to 1/1 (the only way you'll get a zero). To do that you have to apply the exponential power rules.

Therefore,
-6 = 4(x-1)
-6/4 = x-1
-6/4 + 1 = x
-2/4 = x
-1/2 = x

Answer 3

First, convert both sides to the same base, 3/5

(125/27)^(x + 1) = (5^3/3^3)^(x +1) = (5/3)^3(x +1) = (3/5)^(– 3(x
+ 1) = (3/5)^(–3x – 3)

Sustituting the result into the left side of the equation,

(3/5)^(–3x – 3) = (3/5)^(x – 1)

Since exponetial expressions with the same base are only equal if the exponents are equal it is necessary to equate the exponents,

(–3x – 3) = (x – 1), subtract x from both sides and add 3 to both sides,

– 4x = 2, dividing both sides by (–4) gives the final answer,

x = –1/2

Answer 4

let (3/5)^ x= p
try to get as power of 3/5 or 5/3 as (125/27) = (5/3)^3
(125/27)^(x+1) = ((5/3)^3)^(x+1) = p^(-3)(x+1)
= (3/5)^(x-1)
comapring exponetiation

so -3(x+1) = x-1

-3x -3 = x - 1
4x = -2
or x = -1/2

ou can solve for p and then get x

Answer 5

if by this you mean

(125/27)^(x + 1) = (3/5)^(x - 1)
((5^3)/(3^3))^(x + 1) = (3/5)^(x - 1)
((5/3)^3)^(x + 1) = (3/5)^(x - 1)
(5/3)^(3(x + 1)) = (3/5)^(x - 1)
((3/5)^(-1))^(3(x + 1)) = (3/5)^(x - 1)
(3/5)^(-3(x + 1)) = (3/5)^(x - 1)
-3x - 3 = x - 1
-4x = 2
x = (-1/2)