Would you offer me ANY HELP for logarithm/exponent problems?

Question

I’d appreciate it if anyone could offer me any help with ANY/all of these problems with some tips and checking my answer, rather than just giving me the answer! Thanks for any help!

1) Solve for x in the equation

(e^(x+5) / e^(5)) = 3

I found x to equal: (log base e (243)) – 5

OR ln (243) – 5

2) Solve for x in the equation

e^[2 ln x – ln (x^2 + x - 3)] = 1

Would I be able to put natural log signs in front of both sides of the equation and work from there? I had no idea.

3) Solve for x in the equation

3^(2x) – 2*3^(x+5) + 3^10 = 0

I wanted to use the method that, if all of the bases are equal, than you can get rid of them from the equation, but I didn’t know how to get rid of the 2 in the second term.

4) Solve for x in the equation

ln x – ln (x+1) = 1

I got to:

e^1 = x/(x+1)

but I figured that that would be impossible because e^1 is equal to 2. something and the right part of the equation would never be more than one becau

Answer 1

1. e^(x+5) / e^5 = 3, so  e^x = 3, x = ln 3

2. e^[ 2 ln x - ln(x² + x - 3)] = 1

2 ln x - ln(x² + x - 3) = 0

2 ln x = ln(x² + x - 3)

ln x² = ln(x² + x - 3)

x² = x² + x - 3

0 = x - 3

x = 3

3. 3^(2x) - 2•3^(x+5) + 3^10 = 0

3^(2x) + 3^10 = 2•3^(x+5)

divide both sides by 3^(x+5)

3^(x-5) + 3^(5-x) = 2

3^x / 3^5 + 3^5 / 3^x = 2

hmm, a fraction plus its reciprocal = 2, both fractions must be 1, so x = 5

you could also see the equation as quadratic:

(3^x)² - 2•3^5•3^x + (3^5)² = 0

( 3^x - 3^5)² = 0

3^x - 3^5 = 0

3^x = 3^5

x = 5

4.  not impossible, x/(x+1) is a rational function, vertical asymptote at x = -1, and for x < -1, the value of the fraction is > 1.

so e = x/(x+1)

ex + e = x

ex - x = -e

(e-1)x = -e

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