+12 Hi!
Need some help with som complicated logarithm-stuff i don't quite get, and ..
1. a) Solve: logx^3 + 2logx^2 - 14 = 0
b) Solve: (logx)^2 - logx - 6 = 0
2. Write as easy as possible: (i know the answers, but i don't know how to get to them)
a) sqrt(50) / 5 = sqrt(2)
b) (sqrt(24) - sqrt(6))^2 = 6
c) sqrt(27) - sqrt(12) + sqrt(75) - sqrt(18) x sqrt(6) = 0
d) (sqrt(2) - sqrt(3)) x (sqrt(2) + sqrt(3)) = -1
1(a) log x^3 +2logx^2-14=0
Use the fact that log x^n = nlogx, so logx^3 = 3logx and 2logx^2=2(2logx)=4logx to give
3logx +4logx -14=0
7logx =14 logx =2
1(b) (logx)^2-logx-6=0 This is a quadratic equation.Can write like X^2-X-6=0 where X = logx
Factorise to get (X+2)(X-3)=0
So (logx +2)(logx -3) +0 logx=-2 and logx =3
2. Understand that sqrt(ab)= sqrt(a) X sqrt(b) ;that is what these questions are testing.
sqrt(50) = sqrt(25x2) =sqrt(25) X sqrt(2) = 5 sqrt2 so sqrt (50)/5=5(sqrt2)/5 =sqrt2.
All the other examples here are similar,now that you have seen the one I just did,you should have the idea to try the rest
+128597 a) Solve: logx^3 + 2logx^2 - 14 = 0
3log x + 4 log x = 14
7 log x = 14 divide by 7
log x = 2 means that 10^2 = x → x = 100
b) Solve: (log x)^2 - log x - 6 = 0 Factor
[ log x - 3 ] [ log x + 2] = 0 set each factor to 0
log x - 3 = 0 → log x = 3 → 10^3 = x = 1000 .... and........
log x + 2 = 0 → log x = - 2 → 10^-2 = x = 1 / 100
+128597 a) sqrt(50) / 5 = sqrt(2)
√50/ 5 = √ [2 * 25] / 5 = [ √2 *√25] / 5 = [√2 * 5] / 5 = √2 * [5/5] = √2 * 1 = √2
b) (sqrt(24) - sqrt(6))^2 = 6
[√24 - √6] [√24 -√6] =
24 - 2*√24*√6 + 6 =
30 - 2√144 =
30 - 2 (12) =
30 - 24 = 6
c) sqrt(27) - sqrt(12) + sqrt(75) - sqrt(18) x sqrt(6) = 0
[The key here is to wrtie each using a similar sqrt]
√ 27 - √12 +√75 - [√18 * √6]
√[9 * 3] - √[4 *3] + √[25 * 3] - √108
√9* √3 - √4 *√3 + √25 *√3 - √36*√3
3√3 - 2√3 + 5√3 - 6√3 ...... factor out √3 ......
√3 [ 3 - 2 + 5 - 6]
√3 [ 1 + 5 - 6]
√3 [6 - 6 ]
√3 [0] =
0
d) (sqrt(2) - sqrt(3)) x (sqrt(2) + sqrt(3)) = -1
[ √2 -√3 ] [√2 + √3 ]
√2*√2 -√3*√2 +√2*√3 -√3*√3
2 - √6 + √6 - 3
2 - 3 =
- 1
