plz help me with LOG?

Question

1. 4^ (√(X+1)) = 3 ^(3X+2)

2. (log x(<--base) 10) (log 48-log 3)

3. 4 ^(-2log 4 (<-- base)3)

4. a ^(4log a(<--base) x)

5. log (1-x^3)+ log (1+x +x^2) - log (1-x)

6. solve for A
1+ log (AB)= log C

7. solve for B.
A= log 3B - log C

8. 2 log 2(base) (x+2)-log 2(base) (3x-2)=2
show me all your work plz its urgent thank you

Answer 1

1. 4^ (√(X+1)) = 3 ^(3X+2)

Take the log of both sides

(√(X+1)) log 4 = (3X+2) log 3

Square both sides

( X + 1 ) ( log 4 )^2 = ( 9X^2 + 6X + 4 ) ( log 3 )^2

Solve the resulting quadratic equation in X

2. (log x(<--base) 10) (log 48-log 3)

What's the base of the other logs?

3. 4 ^(-2log 4 (<-- base)3)

log X (<-- base) Y is equal to ln X / ln Y so

log 4 (<-- base)3 = ln ( 3 ) / ln ( 4 )

4. a ^(4log a(<--base) x)

Rule of exponents: a ^ ( x y ) = ( a ^ x ) ( a ^ y )
Rule of logs: a ^ ( log_a x ) = x

5. log (1-x^3)+ log (1+x +x^2) - log (1-x)

log ( A B ) = log A + log B
log ( A / B ) = log A - log B

log [ (1-x^3)(1+x +x^2) / (1-x) ]

Simplify the expression in [ ] to complete the problem.

6. solve for A
1+ log (AB)= log C

log (AB) = log C - 1
AB = exp [ log C - 1 ]
A = exp [ log C - 1 ] / B

7. solve for B.
A= log 3B - log C

A + log C = log 3B
A + log C = log 3 + log B
A + log C - log 3 = log B
exp ( A + log C - log 3 ) = B

8. 2 log 2(base) (x+2)-log 2(base) (3x-2)=2

You've seen all the rules requireed, so left as an exercise to the reader :)

Answer 2

assuming all are base 10 2) log_16 x + log_4 x + log _2 x = 7 log(16x)(4x)(2x)=7 128x^3=10^7 x^3=10^7/128 x=(10^7/128)^(a million/3) 3) log _9 x +3 log _3 x = 7 log(9x)(3x)^3=7 243x^4=10^7 x=(10^7/243)^a million/4

Answer 3

1) log[4^(√(X+1))] = log[3^(3X+2)]
(√(X+1)*log4 = (3X+2)*log3
(√(X+1)*log2^2 = (3X+2)*log3
2*(√(X+1)*log2 = (3X+2)*log3

(√(X+1)/(3X+2) = log3/(2log2)


2) logx(10) * (log48 - log3)
= logx(10) * log(48/3)
= logx(10)*log(16)
= logx(10)*log(2^4)
= logx(10)*4log2

Leaving everything in terms of log(base10):

logx(10)*4log2
= (log(10) / logx)*4log2
=(1 / logx)*4log2

= 4log2 / logx

3) log(4^(-2log4(3))
= -2*(log4(3)*log4
= -2*(log3/log4)*log4

= -2log3

4) log( a^4loga(x) )
= 4loga(x) * loga
= 4*(logx/loga)*loga

= 4logx

5) First, we can see that 1-x^3 = (1-x)(1+x+x^2)

So,

log(1-x^3)+ log(1+x +x^2) - log(1-x)
= log(1-x)(1+x+x^2) + log(1+x +x^2) - log(1-x)
= log(1-x) + log(1+x+x^2) + log(1+x +x^2) - log(1-x)
= log(1-x) - log(1-x) + log(1+x+x^2) + log(1+x +x^2)

= 2log(1+x+x^2)

6) logAB = logC - 1
But we can see that 1 = log10 <-- note that log10 is log base 10 of 10

So,

logAB = logC - 1 = logC - log10
logAB = log(C/10)

And, logAB = logA + logB

so,

logA + logB = log(C/10)
logA = log(C/10) - logB
logA = log(C/10B)

And we finally get:

A = C/10B

7) A= log 3B - log C
A = log(3B/C)
10^A = 3B/C
3B = C*10^A

B = C*10^A / 3

8) 2log 2(base) (x+2) - log2(base) (3x-2) = 2
log2(x+2)^2 - log2(3x-2) = 2
log2[ (x+2)^2/(3x-2) ] = 2

Using the change of base formula:

log[ (x+2)^2/(3x-2) ] / log2 = 2
log[ (x+2)^2/(3x-2) ] = 2log2
log[ (x+2)^2/(3x-2) ] = log2^2
log[ (x+2)^2/(3x-2) ] = log4
(x+2)^2/(3x-2) = 4

(x+2)^2 = 4(3x - 2)
x^2 + 4x + 4 = 12x - 8
x^2 - 8x + 12 = 0

Factorizing:

(x - 6)(x - 2) = 0

So the solutions for x are:

x = 2 and x = 6