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

Posted 9 years ago
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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 ?

               

                                                                                              
5 Replies
Thnx guys... i really appreciate your help
In[8]:= Solve[9^(3 x - 2) == 7^(6 x - 4), x] // FullSimplify

During 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 BY: Frank Kampas
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) = 1
Write 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)) = 1
Eliminate 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 Z
Write 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 Z
Isolate 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 Z
Solve 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 BY: Frank Kampas
Thanks mr. Frank Kampas... ;)
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 ...
POSTED BY: Udo Krause
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