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

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

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

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

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.

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.

Remember:

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

Fun stuff you've got there though :)