SOLUTION: Solve for x , show work please. Can someone check my answers please. 1). log4 (3x+4)^5=15 2). log (x^2-9)=log (5x+5) Here is a screenshot of them just incase. Link to sc

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Question 1020039: Solve for x , show work please. Can someone check my answers please.
1). log4 (3x+4)^5=15
2). log (x^2-9)=log (5x+5)
Here is a screenshot of them just incase. Link to screenshot : http://prntscr.com/a42nsg
my answer for #1 was no solution and my answer for #2 was x=7. Thanks !
Found 2 solutions by Cromlix, MathTherapy:
Answer by Cromlix(4381)   (Show Source): You can put this solution on YOUR website!
Hi there,
1) log4 (3x+4)^5=15
(3x + 4)^5 = 4^15
(3x + 4)^5 = 1073741824
3x + 4 = 5√1073741824
3x + 4 = 64
3x = 60
x = 20
2) log (x^2-9)=log (5x+5)
As both sides logs
x^2 - 9 - 5x - 5 = 0
x^2 - 5x - 14 = 0
(x - 7)(x + 2) = 0
x - 7 = 0
x = 7
x + 2 = 0
x = -2
Hope this helps :-)
Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!
Solve for x , show work please. Can someone check my answers please.
1). log4 (3x+4)^5=15
2). log (x^2-9)=log (5x+5)
Here is a screenshot of them just incase. Link to screenshot : http://prntscr.com/a42nsg
my answer for #1 was no solution and my answer for #2 was x=7. Thanks !
1).
NOT QUITE!!



 ------- Applying 

 --------- Dividing by 5



 --------------- Converting to EXPONENTIAL form

3x + 4 = 64

3x = 64 - 4

3x = 60

x = , or 

OR



 







3x = 60

x = , or 



2).


That's right!! Good job!!

log (x^2-9)= log (5x+5)



 ------ Applying 



(x - 7)(x + 2) = 0

x = - 2 is ignored since it's an EXTRANEOUS solution

Therefore,  

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