This equation is damn hard!!!

Algebra Level 4

2^x+2^(x-1)+2^(x-2)=7^x+7^(x-1)+7^(x-2) The number of solutions for x are:

P.s.- if anyone solves it, can you please show how and what are the solutions?


The answer is 1.

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1 solution

Jeffrey Wong
May 28, 2014

2 x + 2 x 1 + 2 x 2 = 7 x + 7 x 1 + 7 x 2 2^x+2^{x-1}+2^{x-2}=7^x+7^{x-1}+7^{x-2}

2 x ( 1 + 1 2 + 1 4 ) = 7 x ( 1 + 1 7 + 1 49 ) 2^x \cdot (1+\frac {1}{2}+\frac {1}{4})=7^x \cdot (1+\frac {1}{7}+\frac {1}{49})

2 x ( 7 4 ) = 7 x ( 57 49 ) 2^x \cdot (\frac {7}{4})=7^x \cdot (\frac {57}{49})

7 × 49 4 × 57 = 7 x 2 x \frac {7 \times 49}{4 \times 57}=\frac {7^x}{2^x}

343 228 = ( 7 2 ) x \frac {343}{228}=(\frac {7}{2})^x

log 343 log 228 = log ( 7 x ) log ( 2 x ) \log343-\log228=\log(7^x)-\log(2^x)

log 343 log 228 = x ( log 7 log 2 ) \log343-\log228=x \cdot (\log7-\log2)

( log 343 log 228 ) / ( log 7 log 2 ) = x (\log343-\log228)/(\log7-\log2)=x

In conclusion: there is only one solution for x

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