since 25=5^2, log(25) = 1/2 log(5) and we have
log(5)x - 1/2 log(5)(x+1) = 0.5
or,
2log(5)x - log(5)(x+10) = 1
log(5)(x^2 / (x+10)) = 1
x^2/(x+10) = 5
x^2 = 5x+50
x^2-5x-50 = 0
(x-10)(x+5) = 0
x=10, since log(-5) is not defined.
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2log(a)x=1+log(a) (7x-10a)
since log(a)a = 1,
log(a) x^2 = log(a) (a(7x-10a))
x^2 = a(7x-10a)
x^2 = 7ax - 10a^2
x^2 - 7ax + 10a^2 = 0
(x-5a)(x-2a) = 0
x = 2a or 5a
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27x3^lgx = 9^1+lg(x-20)
Not sure what this means, but if you meant
27(3^lgx) = 9^(1+lg(x-20))
3^3 * 3^lgx = 3^(2 + 2lg(x-20))
1+lgx = 2+2lg(x-20)
lg10 + lgx = lg100 + lg(x-20)^2
10x = 100(x-20)^2
10x = 100x^2 - 4000x + 400
100x^2 - 4010x + 400 = 0
10(10x-1)(x-40) = 0
x = 1/10 or 40
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log(√a)(1/x)+log(a)x +log(a^2)x +log(a^4)x=c
2log(a)(1/x) + log(a)x + 1/2 log(a)x + 4log(a)x = c
(1/x^2)(x)(√x)(x^4) = a^c
x^(7/2) = a^c
x = a^(2/7 c)
Q1:solve log(5)x-log(25)(x+10) = 0.5
Q2:If 2log(a)x=1+log(a) (7x-10a) ,find x in terms of a.
Q3:Find x for which 27x3^lgx = 9^1+lg(x-20)
Q4:Find x in terms of a and c ,given that log(√ a)(1/x)+log(a)x +log(a^2) x +log (a^4)x=c
6 answers
For the Qn3 ，how u get 3^1+lgx from 3^3 *3^ lgx
？ why not 3^ 3+lgx
？ why not 3^ 3+lgx
And the ans for Qn4 should be x=a^（-4c）...
btw,teacher,how to do this Question? log(5)(5-4x)=log(√5)(2-x) ?
Q4c:
1/x^2*x*sqrt(x)*sqrt(sqrt(x))=a^c;
1/x^(1/4) = a^c
x=a^(-4c)
For:
log5(5-4x)=log√5(2-x)
log5(5-4x)=log5(2-x)²
5-4x=(2-x)²
x=±1
1/x^2*x*sqrt(x)*sqrt(sqrt(x))=a^c;
1/x^(1/4) = a^c
x=a^(-4c)
For:
log5(5-4x)=log√5(2-x)
log5(5-4x)=log5(2-x)²
5-4x=(2-x)²
x=±1
Way to watch, MathMate. I missed the 1/x^2 and read it as x^2. Simpler answer, too.
Note to self: always check your answer!
Note to self: always check your answer!