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

Posted 11 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 ?

               

                                                                                              
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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