solve: log_(4)z+log_4(z-3)=1?

Question

step by step

Answer 1

log_(4)z+log_4(z-3)=1
=>log_(4) [ (z) (z-3) ] = 1 (Using log m + log n = logmn)
=> z (z-3) = 4^ 1 (Using log_m(n) = x => n = m^x)
=> z^2 - 3z =4
=> z^2 - 3z -4 =0
=> z^2 - 4z +z -4 =0
=>z(z-4) +1 (z-4) =0
=>(z+1)(z-4) =0
Thus z = -1 or 4. But since log of negative numbers is not defined. Thus z = only 4.

Answer 2

To solve this homework problem, set
4^LHS = 4^RHS,
where LHS and RHS are the left and right hand sides of the given equation. This yields a quadratic equation for z:
z(z-3) = 4.
The roots are z = 4 and z = -1. You will want to exclude the negative root if you consider the function log_(4)z to be defined only for positive z.

Answer 3

According to the properties of logarithms,
log a + log b= log ab

log 4z+ log 4(z-3) = 1 implies
log 4z(4(z-3)) = 1
i.e log 16z(z-3)=1
log 16(z^2-3z)=1

But we know that log 0= 1

therefore,

16(z^2- 3z)=0

i.e z^2 -3z=0
Z^2=3z
i.e z=3

You can check your answer by substituting in the question

i.e., log 12 + log 4(3-3) = log 12+log 0 = log 12 *0 =log 0 =1

Answer 4

First, a denominator can't be null: (x - 2) ? 0 ? x - 2 ? 0 ? x ? 2 (x - 3) ? 0 ? x - 3 ? 0 ? x ? 3 [one million/(x - 2)] - [one million/(x - 3)] = - one million/2 ? the hassle-free denominator is: (x - 2)(x - 3) [(x - 3)/(x - 2)(x - 3)] - [(x - 2)/(x - 3)(x - 2)] = - one million/2 [(x - 3) - (x - 2)]/[(x - 3)(x - 2)] = - one million/2 (x - 3 - x + 2)/[(x - 3)(x - 2)] = - one million/2 - one million/[(x - 3)(x - 2)] = - one million/2 (x - 3)(x - 2) = 2 x² - 2x - 3x + 6 = 2 x² - 5x + 4 = 0 Polynomial like: ax² + bx + c, the place: a = one million b = - 5 c = 4 ? = b² - 4ac (discriminant) ? = (- 5)² - 4(one million * 4) = 25 - sixteen = 9 = 3² x1 = (- b - ??) / 2a = (5 - 3) / (2 * one million) = one million x2 = (- b + ??) / 2a = (5 + 3) / (2 * one million) = 4 answer : S = { one million ; 4 }

Answer 5

All logs taken as base 4
log (z).(z - 3) = 1
z.(z - 3) = 4^(1) = 4
z² - 3z - 4 = 0
(z - 4).(z + 1) = 0
z = 4 , z = - 1

Answer 6

log_4(z)+log_4(z-3)=1
log_4[z(z-3)]=log_4(4)
z^2-3z=4
z^2-3z-4=0
(z-4)(z+1) = 0
z = 4, -1