inequality with logarithms

[Reecane]

help me to solve
log base 7 (x)<=5+2log base x^1/2 (1/7)

 

[HallsofIvy]

help me to solve
log base 7 (x)<=5+2log base x^1/2 (1/7)

This is \(\displaystyle log_7(x)\le 5+ 2log_{x^{1/2}}(1/7)\)?
First obvious point- this is the same as \(\displaystyle log_7(x)\le 5+ log_{x^{1/2}}(1/49)\).

Use the "change of base" formula: \(\displaystyle log_a(x)= \frac{log_b(x)}{log_b(a)}\).

\(\displaystyle log_{x^{1/2}}(1/49)= \frac{log_7(1/49)}{log_7(x^{1/2})}= \frac{-2}{(1/2)log_7(x)}= -\frac{4}{log_7(x)}\)

So the inequality becomes \(\displaystyle log_7(x)\le 5- \frac{4}{log_7(x)}\). If you let \(\displaystyle y= log_7(x)\) the inequality can be written as \(\displaystyle y\le 5- \frac{4}{y}\).

 

[Reecane]

This is \(\displaystyle log_7(x)\le 5+ 2log_{x^{1/2}}(1/7)\)?
First obvious point- this is the same as \(\displaystyle log_7(x)\le 5+ log_{x^{1/2}}(1/49)\).

Use the "change of base" formula: \(\displaystyle log_a(x)= \frac{log_b(x)}{log_b(a)}\).

\(\displaystyle log_{x^{1/2}}(1/49)= \frac{log_7(1/49)}{log_7(x^{1/2})}= \frac{-2}{(1/2)log_7(x)}= -\frac{4}{log_7(x)}\)

So the inequality becomes \(\displaystyle log_7(x)\le 5- \frac{4}{log_7(x)}\). If you let \(\displaystyle y= log_7(x)\) the inequality can be written as \(\displaystyle y\le 5- \frac{4}{y}\).

Thank you so much. I am getting ready for admission exam in Moscow State University. This is one of the problems in a sample test
It helped a lot. Thank you!

 

[HallsofIvy]

Thank you so much. I am getting ready for admission exam in Moscow State University. This is one of the problems in a sample test
It helped a lot. Thank you!

Moscow, Idaho, U. S. A., not Moscow, Russia?