Logarithmn Help: log_x(4x) - log_4(x) = 4

[shaoen01]

Hi,

Sorry to keep posting questions but i am always stuck halfway. And i would like to thank everyone who has/had helped me!

Qns:
\(\displaystyle \log_{x} (4x) - \log_{4} (x) = 4\)

My Working:
\(\displaystyle \log_{x} (4x) - \log_{4} (x) = 4\)

\(\displaystyle \frac{\log_{2}(4x)}{log_{4}x} + \log_{4}x= 4\)

\(\displaystyle \frac{\log_{2}(4) + \log_{2}x}{log_{2}x} + \log_{4}x= 4\)

But i can't seem to get this answer: x=2,4 --> Does anyone know what i am doing wrong? I tried every way that i can think of but can't seem to get it right. Thanks

 

[stapel]

I don't understand how you're getting your second step...? Or going from that to your third step...?

Try using log rules and the change-of-base formula to convert the first log to base-4:

. . .log_x(4x)

. . . . .= log_x(4) + log_x(x)

. . . . .= [log₄(4)/log₄(x)] + 1

. . . . .= [1/log₄(x)] + 1

Then the equation becomes:

. . . . .[1/log₄(x)] + 1 - log₄(x) = 4

. . . . .1 + log₄(x) - [log₄(x)]² = 4log₄(x)

Can you work with that "quadratic"?

Eliz.