SOLUTION: Hey, I'd be very, very grateful if you could solve the following for me: 5/10^(2x)=7 Thank you so much. :)

SOLUTION: Hey, I'd be very, very grateful if you could solve the following for me: 5/10^(2x)=7 Thank you so much. :)

Algebra.Com

Question 557780: Hey,
I'd be very, very grateful if you could solve the following for me:
5/10^(2x)=7
Thank you so much. :)
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
I'd be very, very grateful if you could solve the following for me:
5/[10^(2x)]=7
10^(2x)=5/7
take log of both sides
(2x)log10=log5-log7
2x=log5-log7
x=(log5-log7)/2≈-0.0731
RELATED QUESTIONS
I'd be very grateful if you could explain me on how to solve this:... (answered by KMST)

I am having trouble trying to solve this problem. I would be very grateful if you could... (answered by ankor@dixie-net.com)

I am having a very hard time proving this identity sin 2x + sin 4x/cos 2x - cos 4x =... (answered by TakeATuition.com)

Could you please help me solve te following equations, I seemto be getting stuck. Thank... (answered by checkley71)

Hi, I need help in solving the following equation. I would be very grateful if you could... (answered by ankor@dixie-net.com)

Hello I am having trouble with this find the rate problem I don't even know where to... (answered by scott8148)

Solve log6 (7) A 6.00 B 0.87 C 1.19 D 2.10 E 1.09 F 2.17 I tried what I... (answered by rapaljer,josmiceli)

could you please help me solve for x in this equation: 2x+y=5 x-y=1 could you by any... (answered by Abbey)

Can you please help me solve this problem? 5/2x+1/4x=7/4+x Thank you so very much! (answered by Alan3354)