if log sqr(x) = 0.7211 find log of x^2?

Answer 1

log (√x) = 0.7211
log (x²) = log ((√x)⁴) = 4 log (√x) = 4*0.7211 = 2.8844

Answer 2

Log x^.5 = .5 log X = .7211: Log X = .7211/.5 = 1.4422
log x^2 = 2 log X= 2* 1.5022 = 2.884

Answer 3

if sqr = sqrt, then here is the solution

log(x^1/2) = 0.7211
using the property of log,

(1/2) * log(x) = 0.7211...so log(x) = 1.4422

now, log(x^2) = 2*log(x) = 2.8844

Answer 4

there's a much easier way to do this problem

rewrite as:
log x^1/2 = 0.7211
which becomes ...
1/2logx = 0.7211

now do the same thing with the other equatiton ...
log x^2
2logx

2 divided by 1/2 = 4

therefore:
0.7211 x 4 = 2.8844

Answer 5

It's 2.8844

You have the log of the square root of x.
If you double it, that's the log of x.
If you double it again, that's the log of x^2.

Answer 6

ok so what you have is log(√x)=.7211
the first step is to get rid of the log
√x=10^.7211
x=(10^.7211)^2=27.682

now all you have to do is take the log of x^2
log((27.682)^2)=log(766.302)=2.8844

Answer 7

I don't understand.

Isn't sqr(x) the same as x^2 ?

So the answer is 0.7211.

Answer 8

It would be 4 time 0.7211 because x^2 = (x^1/2)^4, therefore it's 2.844