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# 9^(3x-2)- 7^(6x-4)

Posted 10 years ago
 Seams that 'step by step' solution for this eqation (for real and imaginary roots)are unavailable in wolfram alpha... Why ?And why are some transcedental functions are 'overkill' for wolfram alpha ?
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Posted 10 years ago
 Thnx guys... i really appreciate your help
Posted 10 years ago
 In[8]:= Solve[9^(3 x - 2) == 7^(6 x - 4), x] // FullSimplifyDuring evaluation of In[8]:= Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. >>Out[8]= {{x -> 2/3}}
Posted 10 years ago
 However, if you move one term to the other side of the equal sign, then you can get a step-by-step solution.Solve for x:9^(3 x-2) = 7^(6 x-4)Look for an exponent to divide both sides by.Divide both sides by 9^(3 x-2):1 = 7^(6 x-4) 9^(2-3 x)Reverse the equality in 1 = 7^(6 x-4) 9^(2-3 x) in order to isolate x to the left hand side.1 = 7^(6 x-4) 9^(2-3 x) is equivalent to 7^(6 x-4) 9^(2-3 x) = 1:7^(6 x-4) 9^(2-3 x) = 1Write exponents in 7^(6 x-4) 9^(2-3 x) = 1 in terms of a common base.7^(6 x-4) 9^(2-3 x) = e^(log(7^(6 x-4))) e^(log(9^(2-3 x))) = e^((6 x-4) log(7)) e^((2-3 x) log(9)) = exp((6 x-4) log(7)+(2-3 x) log(9)):exp(log(7) (6 x-4)+log(9) (2-3 x)) = 1Eliminate the exponential from the left hand side.Take the natural logarithm of both sides:log(7) (6 x-4)+log(9) (2-3 x) = (2 i) pi n  for  n element ZWrite the linear polynomial on the left hand side in standard form.Expand and collect in terms of x:(6 log(7)-3 log(9)) x-4 log(7)+2 log(9) = (2 i) pi n  for  n element ZIsolate terms with x to the left hand side.Subtract 2 log(9)-4 log(7) from both sides:(6 log(7)-3 log(9)) x = 4 log(7)-2 log(9)+(2 i) pi n  for  n element ZSolve for x.Divide both sides by 6 log(7)-3 log(9):Answer: |  | x = (4 log(7))/(6 log(7)-3 log(9))-(2 log(9))/(6 log(7)-3 log(9))+((2 i) pi n)/(6 log(7)-3 log(9))  for  n element Z
Posted 10 years ago
 Thanks mr. Frank Kampas... ;)
Posted 10 years ago
 In[18]:= Reduce[9^(3 x - 2) == 7^(6 x - 4), x]Out[18]= C[1] \[Element] Integers && x == -((2 I \[Pi] C[1] - 4 Log[3] + 4 Log[7])/(6 (Log[3] - Log[7])))In fact it is 9^y==7^(2 y) which of course you see is (3^2)^y == 7^(2 y) which is 3^(2 y)==7^(2 y) which is 3^z ==7^z you can do by hand only ...