Solve for y over the real numbers:
3^(3-y) = 1/9
Take reciporicals of both sides:
3^(y-3) = 9
Multiply both sides by 27:
3^y = 243
243 = 3^5:
3^y = 3^5
Equate exponents of 3 on both sides:
Answer: | y = 5
Sure !
3^3 = 27 in the numerator so you have:
27/3^y = 1/9 multiply both sides by 9
243/3^y = 1 Now multiply both sides by 3^y
243=3^y Now take the log of both sides
log 243 = log (3^y) = y log 3 Now divide both sides by log 3
log 243 / log 3 = y NOw use the calculator to find y = 5
