SOLUTION: Hello there I need help with the following questions please, 1) Log base 5(x^-2x-10)= 2 2) 2^3x-7 = 8 ^2x+2 3) x = log of Square root 512 to the base of square root 2 4

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Hello there I need help with the following questions please, 1) Log base 5(x^-2x-10)= 2 2) 2^3x-7 = 8 ^2x+2 3) x = log of Square root 512 to the base of square root 2 4      Log On


   

Question 473154: Hello there I need help with the following questions please,
1) Log base 5(x^-2x-10)= 2
2) 2^3x-7 = 8 ^2x+2
3) x = log of Square root 512 to the base of square root 2
4) 7^x = 1/343
5) 2x + y^2 - 6y - 12 = 0
any help will be greatly appreciated.
Answer by katealdridge(100) About Me  (Show Source):
You can put this solution on YOUR website!
1) log%285%2C%28x%5E2-2x-10%29%29=+2 Convert this equation to an exponential.
5%5E2=x%5E2-2x-10 Solve this quadratic
25=x%5E2-2x-10
0=x%5E2-2x-35
0=%28x-7%29%28x%2B5%29
x=7x=-5
2) 2%5E%283x-7%29=8%5E%282x%2B2%29 Try to get the bases to be equal. Change 8 to 2%5E3
2%5E%283x-7%29=2%5E%283%282x%2B2%29%29 Distribute the 3
2%5E%283x-7%29=2%5E%286x%2B6%29 Remove bases. Once the bases are the same, the exponents equal each other
3x-7=6x%2B6 Solve for x.
-13=3x
x=-13%2F3
3) x=log%28sqrt%282%29%2Csqrt%28512%29%29 not sure if this is what you meant
Change to an exponential equation
sqrt%282%29%5Ex=sqrt%28512%29 Change 512 to something with a base of 2. Also change square roots into rational exponents
2%5E%28%281%2F2%29%2Ax%29=512%5E%281%2F2%29
2%5E%28%281%2F2%29%2Ax%29=2%5E%289%2A%281%2F2%29%29 Now remove the bases since they are the same.
%281%2F2%29x=9%281%2F2%29 Solve for x.
x=9
4) 7%5Ex+=+1%2F343 Try to get 343 into a base of 7
7%5Ex=343%5E-1
7%5Ex=7%5E%283%2A-1%29
7%5Ex=7%5E-3
x=-3
5) 2x+%2B+y%5E2+-+6y+-+12+=+0+ What are the directions?