log2(log4x) = 1 Solve the

  1. solve the equationlog2(x+4)-log4x=2 the 2 and 4 are lower than the g This is what I got: log2(x+4)+log2(4^x)=2 log2((x+4)*4^x)=2
    1. answers icon 1 answer
    2. anonymous asked by anonymous
    3. views icon 1,053 views
  2. solve the equation.log2(x+4)-log4x=2 please show work
    1. answers icon 1 answer
    2. anonymous asked by anonymous
    3. views icon 363 views
  3. I am having trouble figuring out how to solve these logarithms. Could someone please help!log2(log4x)=1 and solve for x and y:
    1. answers icon 1 answer
    2. Chloe asked by Chloe
    3. views icon 618 views
  4. I am having trouble figuring out how to solve this logarithms. Could someone please help!log2(log4x)=1 and solve for x and y:
    1. answers icon 0 answers
    2. Chloe asked by Chloe
    3. views icon 1,193 views
  5. I am having trouble figuring out how to solve these logarithms. Could someone please help!log2(log4x)=1 and solve for x and y:
    1. answers icon 1 answer
    2. Chloe asked by Chloe
    3. views icon 452 views
  6. log2(log4x) = 1Solve the equation The bases are 2 and 4 respectively, I'm just not sure how to signify that on the keyboard.
    1. answers icon 3 answers
    2. Laura asked by Laura
    3. views icon 1,981 views
  7. i need help with these two homework problemsUse the Laws of Logarithms to combine the expression into a single logarithm log2 5
    1. answers icon 1 answer
    2. Emma asked by Emma
    3. views icon 830 views
  8. 1. Log10²x+log10x²=log10² 2-12. Log4(log2x)+log2(log4x)=2 3. X^logx+5/3= 10^5+log x 4. Log 1/2(x-1)+ log
    1. answers icon 1 answer
    2. karan asked by karan
    3. views icon 1,072 views
  9. solve log2(3x-1)-log2(x-1)=log2(x+1)i have absolutely no idea how to solve this. can anyone help me, please?
    1. answers icon 1 answer
    2. Kelly asked by Kelly
    3. views icon 536 views
  10. Solve:2^(5x-6) = 7 My work: log^(5x-6) = log7 5x - 6(log2) = log7 5x = log7 + 6(log2) x = (log7 + log2^6) / 5 And textbook
    1. answers icon 2 answers
    2. Anonymous asked by Anonymous
    3. views icon 1,230 views