Log5(x) = 16logx(5) solve for x

[TyErd]

log5(x) = 16logx(5)
solve for x.

With this one, I have no idea where to start. All I have even thought about doing is bringing up the 16 to make it 5^16 but that doesn't seem to help me.

 

[Mark44]

Convert one of your logs so that both logs are in the same base. Do you have a formula for converting from one log base to another?

 

[TyErd]

No i don't have a formula to do that

 

[sara_87]

the formula is:

log5(x) = log (x) / log (5)

 

[Mark44]

Let y = log_bx
Then x = b^y
So log x = log(b^y) = y log b
And y = (log x)/(log b)

Hence log_bx = (log x)/(log b)

In the third step above, you can use any log base you want. I used the common log (log₁₀).

 

[TyErd]

ok thnx for the formulas, ok so i have the log base for one side which is (log10(x)) / (log10(5)) what do i do now?

 

[Mark44]

Replace log₅(x) in your original equation.

When you do that, what does your equation become?

 

[TyErd]

(log10(x)) / (log10(5)) = logx(5^16)

 

[Mark44]

ok thnx for the formulas, ok so i have the log base for one side which is (log10(x)) / (log10(5)) what do i do now?

Instead of changing log₅ to log, why don't you change log_x to log₅? The goal is to be using the same log base on both sides of the equation.

 

[TyErd]

so are you saying change it so it is: log5(5^16) / log5(x) = log5(x) ?

 

[Mark44]

Yes. Now put it in the context of the original equation.

log₅x = log_x5¹⁶
==> log₅x = [log₅5¹⁶]/log₅x

The numerator on the right can be simplified to just plain 16, and you can multiply both sides by log₅x.

 

[TyErd]

omg, thankyou so much! i get it finally. I wish i could think like you

 

[Mark44]

Practice...