An algebra problem by Mathoholic Oja
lo g 6 ( x + 3 ) 1 + lo g ( x + 3 ) 2 lo g 0 . 2 5 ( 4 − x ) = 1
Find the positive value of x satisfying the equation above.
Note: This problem was published on page 95 of a book by Prof. AI Prilepko. It was mentioned that the problem was unsolvable but I was able to solve it. I am now looking for better solutions. So please give it a try without guessing.
The answer is 3.
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2 solutions
See when I told not to guess the answer, This is what I told not to guess, You just assumed that both sides are zero, sir. Both the sides could have been 1 or 2 or 3 ...... So I am going to have to disregard your solution. Good attempt though.
lo g 6 ( x + 3 ) 1 + lo g ( x + 3 ) 2 lo g 4 1 ( 4 − x ) lo g ( x + 3 ) lo g 6 − lo g 4 lo g ( x + 3 ) 2 lo g ( 4 − x ) lo g 4 lo g 6 − 2 lo g ( 4 − x ) 2 lo g ( 4 − x ) + lo g 4 lo g ( x + 3 ) − lo g 4 lo g 6 = 1 = 1 = lo g 4 lo g ( x + 3 ) = 0
Solving the equation numerically, we get two solutions x ≈ − 2 . 6 0 6 and x = 3 .
See almost everyone makes it here , What I want is what you did after getting here. It is unsolvable from this step, so you have to use tools such as wolframalpha . I want to see a solution without the use of wolframalpha.
In this case, I don't think it is any algebraic solution. That is what the professor meant. But you problem is not correct because there is two solution.
In India IIT exam we only have to focus on positive integral answers
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Then you have to mention that in the problem. That we have to search for positive integer solution. In that case, since x has to be less than 4 , ( ln ( 4 − x ) must be real) the only allowed values of x are 1 , 2 , 3 , out of which only x = 3 is certainly the solution.
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not necessarily. I never said that x has to be an integer, so there are infinite possibilities, not just 1,2,3. Like 1.1, 2.6, 3.11.....etc.
BTW there is indeed are ways to solve this, one is using geometry and other is more or less an assumption where you assume some things and use basic number theory to get the answer(s)
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Some problems just need to be solved using numerical method. How do we find π , e , and 2 for example? We must well devote more time on something else.
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actually, just for fun, If you want a fun way to calculate pi, just go to my youtube channel, Mathoholic Oja , and watch my video on how archimedes calculated pi.
The given equation can be reduced to the form ln 2 ln ( x + 3 6 ) = ln ( 4 − x ) . One obvious solution arises when both sides of this equation go zero. Fortunately, there is a value of x for this to happen, namely x = 3 ⟹ x + 3 6 = 4 − x = 1 , such that both sides involve ln 1 = 0 .