Asked by Noah
How do I solve the following equation
4-7^2x=1
4-7^2x=1
Answers
Answered by
helper
Assuming you mean
4 - 7^(2x) = 1
7^(2x) = 3
Take logarithm to base 7 of both sides
2x log 7 = log 3
x = log 3/(2 log 7)
x = 0.4771/(2 * 0.8451)
x = 0.4771/1.6902
x = 0.2823
4 - 7^(2x) = 1
7^(2x) = 3
Take logarithm to base 7 of both sides
2x log 7 = log 3
x = log 3/(2 log 7)
x = 0.4771/(2 * 0.8451)
x = 0.4771/1.6902
x = 0.2823
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