SOLUTION: I need help with the following exponential please. Solve: 7^-x-3* 7^1+x -4=0 Please note that -4 is not part of 7^1+x Thank you very much. I have a test in the morning and

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: I need help with the following exponential please. Solve: 7^-x-3* 7^1+x -4=0 Please note that -4 is not part of 7^1+x Thank you very much. I have a test in the morning and      Log On


   

Question 294874: I need help with the following exponential please.
Solve: 7^-x-3* 7^1+x -4=0
Please note that -4 is not part of 7^1+x
Thank you very much. I have a test in the morning and I really want to make sure I have these questions right.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Ugh. Please use more parentheses, they're free and help avoid confusion.
I think this is what you want. If not please re-post.
7%5E%28-x-3%29%2A7%5E%281%2Bx%29-4=0
7%5E%28-x-3%2B1%2Bx%29-4=0
7%5E%28-2%29-4=0
1%2F49=4
Clearly this is not true. There is no solution.
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<<----- ADDITIONAL INFORMATION ----->>
>> This is what I mean. 7%5E%28-x%29-3%2A7%5E%281%2Bx%29-4=0+<<
OK, that's clearer.
+7%5E%28-x%29-3%2A7%5E%281%2Bx%29-4=0
Again, multiply by +7%5E%28x%29.
+1-3%2A7%5E%282x%2B1%29-4%2A7%5E%28x%29=0
Pull out a 7 from the second term.
+1-3%2A7%2A7%5E%282x%29-4%2A7%28x%29=0
Re-arrange.
+-21%2A7%5E%282x%29-4%2A7%5E%28x%29%2B1=0
+21%2A7%5E%282x%29%2B4%2A7%5E%28x%29-1=0
Substitute, let +u=7%5E%28x%29, then +u%5E2=7%5E%282x%29
+21u%5E2%2B4u-1=0
This quadratic is factorable.
+21u%5E2%2B4u-1=%287u-1%29%283u%2B1%29=0
Two solutions using the zero product rule:
+7u-1=0
+u=1%2F7
Now substitute back
+7%5E%28x%29=1%2F7
+x=-1
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+3u%2B1=0
+u=-1%2F3
+7%5E%28x%29=-1%2F3
No solution here, since log requires positive arguments.
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Verify the one solution.
+x=-1
+7%5E%28-x%29-3%2A7%5E%281%2Bx%29-4=0
+7%5E%281%29-3%2A7%5E%280%29-4=0
+7-3-4=0
+0=0
Good solution.
+highlight%28x=-1%29+