2016年5月11日 · x = 7^(-1) = 1/7 The log function is asking us what power of the base gives us the argument of the log, in equation form: If " " x=b^y " " then " " log_b(x) = y The inverse of a log function is a power of the base, i.e. b^(log_b(x)) = b^y= x The way we use this is to apply the "inverse" to both sides of an equation. Starting with the equation from our question log_7x = -1 …

2017年5月17日 · The solution is x in (1,7) Let f(x)=(x-7)/(x-1) The domain of f(x) is D_f(x)=RR-{1} We build a sign chart ...

2018年4月12日 · 8228 views around the world You can reuse this answer ...

2018年3月16日 · I tried this: We can use the definition of log: log_ba=x so that: a=b^x In our case: log_(1/7)x=-1 so that: x=(1/7)^-1=7 using a property of exponents.

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2017年7月27日 · Follow the steps below. The rearranged equation is -13x+5=0, to which the solution is x=5/13. (I'm doing this in more and slower steps than normal to make it super clear.

2018年4月19日 · How do you solve #sqrt(x+7) = x + 1# and find any extraneous solutions? Algebra Radicals and Geometry Connections Radical Equations. 1 Answer

2017年11月20日 · By arranging the equation 7^x = 1/49 = 1/(7^2) (7^x)times(7^2)=1 zeroth power of any number is 1. x+2 = 0 x=-2 This is your answer x=-2

2015年8月10日 · x = 2 Any solution that you will find must satisfy two conditions x+7>=0 implies x>= -7 x+1>=0 implies x>=-1 Overall, the solution(s) to this equation must satisfy the condition x>=-1. Start by squaring both sides of the equation to get rid of the radical term (sqrt(x+7))^2 = (x+1)^2 x+7 = x^2 + 2x + 1 Rearrange this equation into classic quadratic form x^2 + x -6 = 0 …

2015年11月1日 · How do you solve #log_7 7(x+1) + log_7 (x-5)=1#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models. 1 Answer