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 ...

2015年9月1日 · [-7,7] There are two possibilities: Either x is greater than 0, in which case x <= 7 Or, x is less than 0, in which case x >= -7 (because in order for the absolute value of x to be less than 7, x must be greater than -7.) So, x has to be less than or equal to 7, and x has to be greater than -7. So, the solution set will be "from -7 to 7, inclusive". This can be written like this: [-7, 7]

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-axis at 2 different values of x. Therefore there are 2 roots. graph{2x^2+5x-7 [-20, 20, -12,12] [-20, 20, -12, 12]} The give equation is 2x^2+5x-7=0 By factoring method, 2x^2+5x-7=0 (2x+7)(x-1)=0 by the …

2016年7月18日 · x = 7/6 When working with logs, the terms must all be in either log form or as numbers, but not a combination of both. note: log_10 10 = 1 and log_2 2 = 1 and log_7 7 =1 etc. In order to have all the terms written as logs with the same base, the given equation can be changed to log_7x-log_7(x-1)=log_7 7 If log terms are added, they can be written as the log of …

2018年5月8日 · From the coefficients, you can tell that #x=1# will make the equation equal to zero. #2(1)^3 -7(1)^2 -2(1) +7 = 0# Then you do a polynomial divsion of #(2x^3-7x^2-2x+7)/(x+1)# (Don't know how to format this so I wrote it out and took a picture of it) #2x^3-7x^2-2x+7 = (x+1)(2x^2-9x+7)#

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 …

2018年4月16日 · 7y+12x=-16 Standard form is ay+bx=c Start by multiplying everything by 7 to get rid of pesky fraction: 7y+28=-12(x-1) Distribute: 7y+28=-12x+12 Than subtract 28 over and add the 12x over: 7y+12x=-16