Answer:
[tex]x = \frac{3ln(3)+ln(5)}{ln(3)-4ln(5)}[/tex]
Step-by-step explanation:
1) Since we cannot cancel out the bases, take the natural log of both sides.
[tex]3^x^-^3 = 5^4^x^+^1\\ln 3^x^-^3 = ln 5^4^x^+^1[/tex]
2) Now, use the power property of logarithms to move the exponents out in front like so:
[tex]ln 3^x^-^3 = ln 5^4^x^+^1\\(x-3)ln3 = (4x + 1) ln5[/tex]
3) Now, expand out the terms inside the parentheses:
[tex](x-3)ln3 = (4x + 1) ln5\\xln3-3ln3=4xln5+ln5[/tex]
4) Now, move all the terms with the x variable onto one side:
[tex]xln3-3ln3=4xln5+ln5\\xln3-4xln5 = 3ln3+ln5[/tex]
4) Now, factor the x out of the x terms on the left side:
[tex]xln3-4xln5 = 3ln3+ln5\\x(ln3-4ln5)=3ln3+ln5[/tex]
5) Finally, divide both sides by [tex]ln3-4ln5[/tex] to isolate and solve for x.
[tex]x(ln3-4ln5)=3ln3+ln5\\x = \frac{3ln3+ln5}{ln3-4ln5}[/tex]
Thus, x = [tex]\frac{3ln(3)+ln(5)}{ln(3)-4ln(5)}[/tex].
