+0  
 
+3
78
2
avatar+598 

Solve and find the domain of the equation:

 

\(\log _3 (4 \cdot 3^{x-1}-1)=2x-1\)

michaelcai  Feb 12, 2018
 #2
avatar+86944 
+1

log3 (4*3x- 1  - 1)   = 2x - 1   write this in exponential form

 

3^(2x - 1)  =  4*3^(x - 1) - 1

 

3^(2x) / 3  =  4*3^x / 3  - 3/3       multiply through by 3

 

3^(2x)  = 4*3^x  -  3

 

(3^2)^x  = 4*3^x - 3

 

(3^x)^2  = 4*(3^x)  -  3

 

Let  a  =  3^x

 

a^2  =  4a  - 3

 

a^2  - 4a + 3   =  0       factor

 

(a - 3)  (a - 1)  =  0

 

Set  each factor to 0  and solve for a and we get that

 

a  =  3    or    a  =1        which means that

 

1 =  3^x    ⇒  x = 0      or  that

 

3  =  3^x   ⇒   x  =  1

 

Both solutions solve the original equation

 

I'm sorry...but I don'tknow what "domain of the equation"  means....unless we require that 

 

4*3^(x - 1)  -  1  >  0    if so.....then

 

4*3^(x - 1)  .  1

 

3^(x - 1)  > 1/4     take the log of both sides

 

(x - 1) log(3)  > log (1/4)

 

log (3) *  (x)  >  log (1/4) + log (3)

 

x > [ log (1/4) + log (3) ] / log (3)  

 

x >  log (3/4) / log(3)

 

x > ≈  -0.262  ⇒    this is the domain

 

 

cool cool cool

CPhill  Feb 13, 2018
edited by CPhill  Feb 13, 2018
edited by CPhill  Feb 13, 2018

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