log (6x + 5) = 26x + 5 > 06x + 5 - 5 > 0 - 56x > - 56x/6 > -5/6x > -5/6log (6x + 5) = 210^2 = 6x + 5100 = 6x + 5100 - 5 = 6x + 5 - 595 = 6x95/6 = 6x/695/6 = xCheck:
2a+1
2logx-log5=-2 logx^2-log5=-2 log(x^2 / 5)=-2 x^2 / 5 = 10^-2 x^2=5(1/100) x^2=1/20 x=√(1/20)
12.6
2 log(x) + 3 log(x) = 105 log(x) = 10log(x) = 10/5 = 210log(x) = (10)2x = 100
No. log 20 is a positive number , so it you subtract it from log 5 you get less than log 5. However, log10 5 = 1 - log102 = 2- log1020 . or log 5 - log 20 = log 5 - log 4*5 = log 5 - (log 5 + log 4) = log 5 - log 5 - log 4 = - log 4 But we do not need to do all of these computations, because log 5 is different from log 5 - log 20 by the law of the equality that says two equals remain equal if and only if we subtract (in our case) the same thing from them.
Adam-12 - 1968 Log 63 Baby 2-7 was released on: USA: 8 November 1969
63 x 5 = 315
5/9 + 5/7 = 35/63 + 45/63 = 80/63 or 1 20/63
7x = 5x log(7) = log(5)x = log(5) / (log(7) = 0.82709 (rounded)
45/63 45 ÷ 9 = 5 63 ÷ 9 = 7 so 45/63 = 5/7; 5 to 7; 5:7
[log2 (x - 3)](log2 5) = 2log2 10 log2 (x - 3) = 2log2 10/log2 5 log2 (x - 3) = 2(log 10/log 2)/(log5/log 2) log2 (x - 3) = 2(log 10/log 5) log2 (x - 3) = 2(1/log 5) log2 (x - 3) = 2/log 5 x - 3 = 22/log x = 3 + 22/log 5
100000 log = 5.
63
log (6x + 5) = 26x + 5 > 06x + 5 - 5 > 0 - 56x > - 56x/6 > -5/6x > -5/6log (6x + 5) = 210^2 = 6x + 5100 = 6x + 5100 - 5 = 6x + 5 - 595 = 6x95/6 = 6x/695/6 = xCheck:
It is: 58+5 = 63
5 / 7 + 5 / 9 lowest common denominator = 63, so 45 / 63 + 35 / 63 = 80 / 63 = one and seventeen sixty thirds ( 117/63