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 !
<b><u>1).</b></u>
<pre>NOT QUITE!!
{{{log (4, (3x + 4))^5 = 15}}}
{{{5 * log (4, (3x + 4)) = 15}}} ------- Applying {{{log (a, b)^c = c * log (a, 
b)}}}
{{{log (4, (3x + 4)) = 15/5}}} --------- Dividing by 5
{{{log (4, (3x + 4)) = 3}}}
{{{3x + 4 = 4^3}}} --------------- Converting to EXPONENTIAL form
3x + 4 = 64
3x = 64 - 4
3x = 60
x = {{{60/3}}}, or {{{highlight_green(x = 20)}}}
OR
{{{log (4, (3x + 4))^5 = 15}}}
{{{(3x + 4)^5 = 4^15}}} 
{{{(3x + 4)^5 = (4^3)^5}}}
{{{3x + 4 = 4^3}}}
{{{3x = 4^3 - 4}}}
3x = 60
x = {{{60/3}}}, or {{{highlight_green(x = 20)}}}

<b><u>2).</b></u>
<pre>That's right!! Good job!!
log (x^2-9)= log (5x+5)
{{{log ((x^2 - 9)) = log ((5x + 5))}}}
{{{x^2 - 9 = 5x + 5}}} ------ Applying {{{log (a) = log (b)}}}
{{{x^2 - 5x - 14 = 0}}}
(x - 7)(x + 2) = 0
x = - 2 is ignored since it's an EXTRANEOUS solution
Therefore, {{{highlight_green(x = 7)}}}