Logarithms help about showing changes in logs?

Question

Still running into problems with logarithms!

If log*base 2* (x) = a

Then how would I show that log*base 2* (16x) = 4+a ?

And that log*base 2* (x^4 / 2) = 4a-1

Answer 1

log_2(x) = a

log_2(16x) = log_2(16) + log_2(x) = 4 + a

log_2(x^4 / 2) = log_2(x^4) - log_2(2) = 4 * log_2(x) - 1 = 4a - 1

Answer 2

Let start with the first:

If log*base 2* (x) = a, then in exponential form, you'd have...

2^a = x

Multiply both sides by 16...

16(2^a) = 16x

but 16 = 2^4, so...

(2^4)(2^a) = 16x

When you mulitply powers with the same bases, you add the exponents, so...

(2^ (4+a)) = 16x

Rewrite back in exponential form, you have...

log*base 2*(16x) = 4+a

Part 2 coming soon to a theater near you.

Answer 3

log 16x base a
=log 16 base 2+log x base 2
=4log 2 base 2+a
=4+a

log x^4/2 base 2
=log x^4base 2-log 2 base2
4log x base2-1
=4a-1

Answer 4

let´s log mean logarithm in base 2

log16x= log 2^4 x = 4log2 +logx.But log2=1 So It is 4+a

log x^4/2 = 4logx -log2 = 4a-1