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Solve for x. log2(7x + 1) = log2(2 - x) a) 1/8 b) 8 c) 1/4 d) 4
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7x + 1 = 2 - x solve for x
Since the two logs are to the same base and there are no other terms, the expressions inside the logarithms must be equal. So, we can take 2 and raise it to the power of each side to get: \[7x+1=2-x\]From here you can solve for x.
7x+1=2-x -1 -1 7x=1-x +x +x 7x^2=1 x^2=1/7 im so confused !??
7x + x = 8x
Simply take antilog on base 2 on both sides and you will get what RadEn posted in the first reply.
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