What is The true solution to In 20 +in 5=2in x?

  x = 10

Step-by-step explanation:

Combining the logarithms, you have ...

 ln(20·5) = ln(x²)

  100 = x² . . . . . take the antilogs

  10 = x . . . . . . . take the positive square root

Alternate solution

  ln(2·2·5) +ln(5) = 2·ln(x) . . . . . . . factor 20

  2ln(2) +ln(5) +ln(5) = 2·ln(x) . . . . rearrange ln(20)

  ln(2) +ln(5) = ln(x) . . . . . . . . . . . . divide by 2

  ln(10) = ln(x) . . . . . . . . . . . . . . . . combine logs on the left

  10 = x

These solutions make use of the rules of logarithms ...

  ln(ab) = ln(a) +ln(b)

  ln(a²) = 2·ln(a) . . . . . same as the first rule for a=b

step-by-step explanation:

you can first factor out the n to get n(1+5). we know that 1+5 is 6, so we just have 6n.