Solve log2 8 + log232.?

Question

Solve log2 8 + log232

Answer 1

log2 8 is log8/log2
log2 32 is log32/log2

now just plug it in a calculator!
you should get 8.

if you can't use a calculator, set it up like this:
2^x=8 and 2^y=32
then add x and y (3 and 5)

Answer 2

using the change of base formaula
5.365487985 hoping that its log232 and not log base 2 of 32 if it is that then
8

Answer 3

plug this into ur calc (log 8 / log 2) + (log 32 / log2)

Answer 4

you may desire to remember the definition of a log. If log(base b)(x) = a, then b^a = x So, you are able to convert your equation right into a go browsing the two factors fairly actual, via remembering that log2(2^-3) = -3 log2 (2x+8) - log2 (2x^2 + sixty one) = -3 log2 (2x+8) - log2 (2x^2 + sixty one) = log2(2^-3) log2[(2x+8)/(2x^2 + sixty one)] = log2(2^-3) (2x+8)/(2x^2 + sixty one) = 2^-3 2(x+4)/(2x^2+sixty one) = one million/8 (x+4)/(2x^2+sixty one) = one million/sixteen x+4 = (2x^2+sixty one)/sixteen 16x + sixty 4 = 2x^2 + sixty one 2x^2 - 16x -3 = 0 x = [sixteen ±?(sixteen^2 - 4*2*-3)/4 = [sixteen±?280]/4 = [sixteen±2?70]/4 = (8±?70)/2