×

# Stuck!

Here is a problem from the NMTC 2015 Bhaskara Contest:

$$\text{The number of values of}\space x\space \text{which satisfy the equation}\space 5^2.\sqrt[x]{8^{x-1}}=500\space \text{are:}$$

Here is the link. (Question 6).

I solved it like this:

$$5^2.\sqrt[x]{8^{x-1}}=500$$

$$\Rightarrow\space 5^2.(8^{x-1})^{\frac{1}{x}}=2^2\times5^3$$

$$\Rightarrow\space 5^2.8^{\frac{x-1}{x}}=(2^3)^{\frac{2}{3}}\times5^3$$

$$\Rightarrow\space 8^{\frac{x-1}{x}}=8^{\frac{2}{3}}\times5$$

But here, I am stuck. Any idea?

Note by Ashok Dargar
1 month, 1 week ago

Sort by:

Actually, this is my father's account, and now I use it. I am in 7th standard. So please can you explain what 'log' is or explain without log?😊 · 1 month, 1 week ago

Check out the Brilliant wiki to learn logarithms which is pretty good. · 1 month, 1 week ago

There is no solution of this equation. if have any give that soln · 2 weeks, 4 days ago

It has no integral but a real sol · 2 weeks, 4 days ago

Yep you are right buddy! It has no integral solutions but one real solution. So the answer is 1 as the question does not specify whether they should be integers or real number. · 2 weeks, 3 days ago

Sir, you can continue from your approach by dividing both sides by $$8^{\frac{2}{3}}$$ (in your final equation) and the applying log to the base $$10$$ on both sides(A one step (log rules) simplification is required). This equation is obviously a linear equation, thus having only one solution. · 1 month, 1 week ago