Solve the equation (1÷49)^x + 2 =(7)^x - 3​

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  x = -1/3

Step-by-step explanation:

Maybe you mean ...

  (1/49)^(x +2) = 7^(x -3)

Taking logarithms base 7, we have ...

  -2(x +2) = x -3

  -2x -4 = x -3 . . . . . eliminate parentheses

  -1 = 3x . . . . . . . . . . add 3+2x

  x = -1/3 . . . . . . . . . divide by the coefficient of x

This is confirmed by a calculator in the first attachment.

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As written, the equation is ...

  ((1/49)^x) +2 = (7^x) -3

Multiplying both sides by 7^(2x) gives us ...

  1 + 2·7^(2x) = 7^(3x) -3·7^(2x)

Subtracting the left side gives us the cubic equation in 7^x:

  (7^x)^3 -5·(7^x)^2  -1 = 0

This has one irrational real root (found using a graphing calculator). The solution is ...

  x ≈ 0.83111881051

This answer is confirmed in the second attachment. (10^-16 is essentially zero for this purpose.)

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Additional comment

Part of the reason for going to the trouble to solve the "as written" equation is to show you that it is possible, though messy. What we'd like to emphasize here is that grouping symbols are needed when exponents contain math operations.

In typeset text, the superscript font indicates the grouping of parts of an exponent. In plain text, we cannot tell if 7^x-3 means (7^x)-3 or 7^(x-3). The Order of Operations tells us to interpret it the first way: (7^x) -3, because exponentiation is done before addition. That leads to trouble in this problem.

start with a line segment pq that we will copy. mark a point r that will be one endpoint of the new line segment. set the compasses' point on the point p of the line segment to be copied. adjust the compasses' width to the point q. the compasses' width is now equal to the length of the line segment pq.

step-by-step explanation: