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 …

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 ...

2017年1月17日 · {36/7} Recall that x^2 - 5x can be written as x(x - 5). 1/(x(x - 5)) = (x + 7)/x - 1 Put on a common denominator: 1/(x(x -5)) = ((x + 7)(x - 5))/(x(x - 5)) - (x(x -5 ...

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.

2018年3月2日 · The roots are x=-7/2 and x=1 graph{2x^2+5x-7 [-20, 20, -12,12] [-20, 20, -12, 12]} One way to find the number of roots is by the graph. It is clear that the graph crosses the x …

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 …

2017年6月17日 · Now, we have a quadratic equation. We can factorise this to (x+5)(x-7) =0 Set each factor equal to 0, since if either (x+5) or (x-7) was 0, the equation would hold, because 0 …

\[7x-1=1\] +. > < ...

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