Exponential equation

[sashary]

The problem is "Find the exact solution to 3^x = 5^(x-1)"

I have tried to take the natural log of both sides x ln(3) = x-1 ln(5) but I'm stuck on what to do next.

 

[mmm4444bot]

x ln(3) = x - 1 ln(5)

This is wrong, without grouping symbols to change the Order of Operations.

x ln(3) = (x - 1) ln(5)

Now use the Distributive Property, to expand the righthand side.

Then solve for x, as in any other linear equation. Get all of the terms that contain x to one side; get all of the constant terms to the other side. Factor out x, and isolate it.

For comparison, I get 3.1507 (rounded) as the decimal equivalent for my resulting expression for x.

Thank you for showing your work, thus far.

 

[sashary]

This is what I'm getting now.

x ln(3) = (x - 1) ln(5)

x ln(3) = x ln(5) -ln(5)

x (ln (3) -ln(5)) = -ln(5)

x = -ln(5)/(ln(3) -ln(5))

 

[Denis]

You lost an "x" ... go find it... :shock:

 

[mmm4444bot]

This is what I'm getting now.

x = -ln(5)/(ln(3) - ln(5))

Your latest work looks okay (to me), and your expression for x is correct, but it can be simplified.

If we factor out -1 from the denominator, it will cancel with the factor of -1 in the numerator.

Doing that yields:

ln(5)/[ln(5) - ln(3)]

Next, there is a property of logarithms that allows us to express the denominator above as a single logarithm.

Doing that yields:

ln(5)/ln(5/3)