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# C2 log question

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1. what will happen if I times log2X to log2X? Will it be (log2X)2? If it is, how do i solve this?
2. You're correct.
You should get 2 values of x for this equation.
3. Notice 32=2^5 and 16=2^4. Use laws of logs... And then use part a
4. The values of are just integers so the numerator simplifies very nicely.
5. log2 32 + log2 16 means you can multiply them. So its log2 515 / log2 x = log2 x

multiply x to both sides

log2 512 = (log2 x)^2

if you type log2 512 into your calculator, that will be simplified down to 9

sq root both sides

+/- 3 = log2 x

using log rules, this is saying x is equal to 2 to the power of plus or minus three.

so x = 8 or 1/3

sorry, its hard to explain logs without actually writing it out
6. (Original post by Mina_)
log2 32 + log2 16 means you can multiply them. So its log2 515 / log2 x = log2 x

multiply x to both sides

log2 512 = (log2 x)^2

if you type log2 512 into your calculator, that will be simplified down to 9

sq root both sides

+/- 3 = log2 x

using log rules, this is saying x is equal to 2 to the power of plus or minus three.

so x = 8 or 1/3

sorry, its hard to explain logs without actually writing it out
There's no need to multiply. If you notice 32=2^5 and 16=2^4 then you can evaluate the top without a calculator as 9. So 9/y=y, y^2=9 , y=±3 where y=log_2(x) as you said...

Answers should be 8 and 1/8.
7. (Original post by alesha98)

what will happen if I times log2X to log2X? Will it be (log2X)2? If it is, how do i solve this?
you can solve

you can also sub if you find doing things with algebra help
8. (Original post by Math12345)
There's no need to multiply. If you notice 32=2^5 and 16=2^4 then you can evaluate the top without a calculator as 9. So 9/y=y, y^2=9 , y=±3 where y=log_2(x) as you said...

Answers should be 8 and 1/8.
Sorry, I meant 1/8 !
9. (Original post by Mina_)
Sorry, I meant 1/8 !
Thanks, it is such an easy question but i overthinking it

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