I am given it a try concerning the conversion;

Log x = 1/2log x

so, it will be

log (x) = 1/2log (x)

Is that correct?

]]>Sorry, I was away pretty much all day. Where are you stuck?

Looks like the first thing you should do is convert those weird logs to the natural log and then you should have no problem.

Do you know how?

]]>I don't understand the question.

]]>Log x = 1/2logx and hence solve log x + 1/2 = logx

In both cases the LHS is in base 4 and the RHS is in base 2

Thanks

]]>Also that is not the equation you gave in post #664. In post #664 you gave this equation

log(x+1) - 2logx^2=1

That is the equation I said does not have a root of 1 / 3.

Now you are giving this equation.

EbenezerSon wrote:

This is it;

log(x^2 + 1) - 2logx = 1

They are not the same.

]]>log(x^2 + 1) - 2logx = 1

]]>Ask the tutor to plug and check his answer.

Yes - have seen the tutor - the tutor gave me a calculator which I plugged 1/3 into the equation and I had 1(one), so it seems his answer is correct. What do you say Bobbym?

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