Logarithm

[Matthew Prey]

What Value of the root product ?

 

[tkhunny]

What is the actual question?

Perhaps think on this: $$125\cdot x^{Something} = x^{5}$$
Is it interesting that we have $$\log_{5}()\;and\;125 = 5^{3}$$
What are your thoughts? What have you tried? Explore the possibilities!

 

[Dr.Peterson]

Does "root product" mean "product of all roots"?

My first thought is to simplify the problem by making a substitution, perhaps [MATH]u = log_5 x[/MATH]. Then you will be able to solve for u.

 

[Matthew Prey]

Does "root product" mean "product of all roots"?

My first thought is to simplify the problem by making a substitution, perhaps [MATH]u = log_5 x[/MATH]. Then you will be able to solve for u.

Sorry, is "product of all roots"

 

[JeffM]

What Value of the root product ?

View attachment 12638

When I see an unknown in an exponent, I think that taking logs MAY well be helpful.

[MATH]\text {Given } a, \ b > 0,\ a = b \iff log_c(a) = log_c(b).[/MATH]
Moreover, I see a log to the base 5 on one side of the equation and a power of 5 on the other side of the equation. So I EXPERIMENT by using logs to the base 5. Algebra gives you a bag of tools. If you are not sure which will help, try one that seems reasonable.

[MATH]x^{log_5(x^2)} = \dfrac{x^5}{125} \implies log_5(x^{log_5(x^2)}) = log \left ( \dfrac{x^5}{125} \right ).[/MATH]
This LOOKS ugly, but maybe simplification will help. Use the laws of logs to do so.

[MATH]\therefore log_5(x^2) * log_5(x) = log_5(x^5) - log_5(125) \implies 2 * log_5(x) * log_5(x) = 5log_5(x) - 3.[/MATH]
Now what, or is this a dead end?

 

[Matthew Prey]

When I see an unknown in an exponent, I think that taking logs MAY well be helpful.

[MATH]\text {Given } a, \ b > 0,\ a = b \iff log_c(a) = log_c(b).[/MATH]
Moreover, I see a log to the base 5 on one side of the equation and a power of 5 on the other side of the equation. So I EXPERIMENT by using logs to the base 5. Algebra gives you a bag of tools. If you are not sure which will help, try one that seems reasonable.

[MATH]x^{log_5(x^2)} = \dfrac{x^5}{125} \implies log_5(x^{log_5(x^2)}) = log \left ( \dfrac{x^5}{125} \right ).[/MATH]
This LOOKS ugly, but maybe simplification will help. Use the laws of logs to do so.

[MATH]\therefore log_5(x^2) * log_5(x) = log_5(x^5) - log_5(125) \implies 2 * log_5(x) * log_5(x) = 5log_5(x) - 3.[/MATH]
Now what, or is this a dead end?

Excellent !!! Thank you very much !!