Maths help: algebra with logarithms?

So, since logarithms are opposite of exponents, you will need to use log base e (or natural log, abbreviated ln) here.

We need some initial values, so let's substitute 65 for x (initial), and 41 for y (ultimatley what we want).

So, now you have 41 = 65e^(-65t) and we're solving for t. First off, divide both sides by 65, and we're now left with: 41/65 = e^(-65t). From here, take the natural log of both sides, and you'll get:

ln(41/65) = -65t and you can solve from there.

Substituting the figures in the question gives:

41 = 65e^(-450k)

e^(-450k) = 41/65 → e^450k = 65/41

Converting this into a natural log expression :

450kLn(e) = 450k = Ln 1.585366 = 0.460815

k = 0.460815/450 = 0.001024 years .........As this is approximately 9 hours either I am wrong or the model formula is incorrect. Please check, as the expected answer is 300-350 years

log (2x + a million) = a million/3log (x+a million) shop on with the regulations log (2x + a million) = log (x+a million)^a million/3 2x+a million = (x +a million)^a million/3 cube the two facets (2x+a million)^3 = x+a million 8x^3 + 12x^2 + 6x + a million = x + a million transpose x and +a million to the different area making it -x and -a million 8x^3 + 12x^2 + 5x = 0 ingredient out x x(8x^2 + 12x + 5) = 0 x = 0 8x^2 + 12x + 5 = 0 use quadratic formulation for this one a = 8 b = 12 c =5 x = { -b +/- [b^2 - 4ac]^a million/2}/2a x = { -12 +/- [12^2 - 4(8)(5)]^a million/2}/2(8) x = { -12 +/- [one hundred forty four - a hundred and sixty]^a million/2}/sixteen x = { -12 +/- [-sixteen]^a million/2}/sixteen x = { -12 +/- 4i}/sixteen x = { -3 +/- i}/4 --it truly isn't anymore seen a real answer it truly is a(n) imaginary root(s) via presence of i

Surely you have to substitute (t) into the equation as well? therefore you will have:

ln(41/65)= -450*k