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endlesslove230 · Oct 15, 2011, 01:08 AM

$5^{(log_5)^2 {x}} x^{log_5 x} < 10$

$(sqrt{2} 1)^{\frac{6(x-1)}{x 1}} \leq (sqrt{2} - 1)^{-x}$

endlesslove230 · Oct 15, 2011, 01:10 AM

$5^{(log_5)^2 x} + x^{log_5 x} < 10$ sorry...

endlesslove230 · Oct 15, 2011, 01:12 AM

$(sqrt{2} + 1)^{\frac{6(x-1)}{x+1}} \leq (sqrt(2) - 1)^{-x}$

ebaines · Oct 15, 2011, 06:36 AM

Originally Posted by endlesslove230

$5^{(log_5)^2 x} + x^{log_5 x} < 10$ sorry...

You need to be careful with the parentheses. For the first part of this equation you wrote:

$5^{(log_5)^2 x} $

which doesn't make sense. Did you mean:

$ 5^{(log_5 x)^2} $
or perhaps
$ 5^{(log_5 x^2)} $ ?

In any event, you should be able to solve this if you apply the principle that
$ a ^{log_a b} = b $

Post back what you get and we'll check it for you.

endlesslove230 · Oct 15, 2011, 12:17 PM

In fact I mean that 2 is the power of the logarithm, not of x or 5.

I know that formula, but I'm not sure if I can apply it in this case.

If I had $(5^{log_5 x})^2$ , the result was clearly $x^2$ .

But I have $5^{{log_5}^2 x} = 5^{{log_5 x}{log_5 x}}$ , so I don't think I can use it any more.

Unknown008 · Oct 16, 2011, 07:02 AM

Remember:

$a^{mn} = (a^m)^n$

Fun stuff you've got there though :)