maths explain please

r035152000

2012-07-09 9:20 am

Q1
if log5=p and log4=q, find log 200
A. 2(q+p)
B. 2p=(3/2)q
C. p^2q^(3/2)
D. p^2+q^(3/2)

Q2
if 5^x =7^y=10^z and x,y,z are non-zero, then z{(1/x)+(1/y)}=
A.24/7
B.10/35
C.log35
D.log70

Q3
solve 9^x-3^(x+1)=40
A.1.2
B.1.6
C.1.9
D.2.1

Q4
solve log base 3 (x+3)=log base 9 (3x+19)
A. -2/3
B.2
C.7/2
D.5

回答 (1)

50418129

2012-07-09 6:04 pm

✔ 最佳答案

(1) B

log 200
= log (5^2 x 2^3)
= log (5^2 x 4^(3/2))
= 2 log 5 x (3/2) log 4
= 2p + (3/2)q

(2) C

let k = 5^x =7^y=10^z
so log k = x log 5 = y log 7 = z

thus z[(1/x) + (1/y)]
= log k[(log 5 / log k) + (log 7 / log k)]
= log 5 + log 7
= log 35

(3) C

9^x - 3^(x + 1) = 40
(3^x)^2 - 3(3^x) - 40 = 0
[(3^x) - 8][(3^x) + 5] = 0
3^x = 8 or 3^x = -5 (rejected)
x = log 8 / log 3 = 1.9

(4) B

log base 3 (x + 3) = log base 9 (3x + 19)
log (x + 3) / log 3 = log (3x + 19) / log 9
log (x + 3) / log 3 = log (3x + 19) / 2 log 3
2log (x + 3) = log (3x + 19)
(x + 3)^2 = 3x + 19
x^2 + 3x -10 = 0
x = 2

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