Can someone point me in the right direction how to solve this?
$3^{x+2}\cdot4^{-(x+3)}+3^{x+4}\cdot4^{-(x+3)} = \frac{40}{9}$
I guess I have to get to logarithms of the same base. But how? What principle should I use here? Thx
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1Hint: factor out the fours and threes, its much easier then. – Ziad Fakhoury Jun 16 '16 at 13:35
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$3^{x+4}=3^{2}\cdot 3^{x+2}$ – Surb Jun 16 '16 at 13:36
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See https://proofwiki.org/wiki/Exponent_Combination_Laws – lab bhattacharjee Jun 16 '16 at 15:45
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$$3^{ x+2 }\cdot 4^{ -(x+3) }+3^{ x+4 }\cdot 4^{ -(x+3) }=\frac { 40 }{ 9 } \\ { 4 }^{ -\left( x+3 \right) }{ 3 }^{ x+2 }\left( 1+9 \right) =\frac { 40 }{ 9 } \\ { 4 }^{ -\left( x+3 \right) }{ 3 }^{ x+2 }=\frac { 4 }{ 9 } \\ { \left( \frac { 3 }{ 4 } \right) }^{ x }\frac { 9 }{ 64 } =\frac { 4 }{ 9 } \\ { \left( \frac { 3 }{ 4 } \right) }^{ x }={ \left( \frac { 4 }{ 3 } \right) }^{ 4 }\\ x=-4\\ \\ $$
haqnatural
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2I am not the downvoter, but i believe he may have downvoted because you provided the solution when OP asked for guidance. – Ziad Fakhoury Jun 16 '16 at 13:40
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That transition from 3rd to 4th line is was I was missing. Thank you, sir! – Smejki Jun 16 '16 at 13:49
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Using $a^{-b}=\frac{1}{a^b}$, The equation can be rewritten as $$\frac{3^{x+2}}{4^{x+3}}+\frac{3^{x+4}}{4^{x+3}}=\frac{40}{9}$$ $$\implies \frac{9}{64}\left(\frac{3}{4}\right)^x+\frac{81}{64}\left(\frac{3}{4}\right)^x=\frac{40}{9}$$ $$\implies \left(\frac{3}{4}\right)^x=\frac{256}{81}$$ $$\implies x=-4$$
Nikunj
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1I am not the downvoter, but i believe he may have downvoted because you provided the solution when OP asked for guidance. – Ziad Fakhoury Jun 16 '16 at 13:41
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A hint to get you started:
$$3^{x+2}\cdot 4^{-(x+3)}+3^{x+4}\cdot 4^{-(x+3)}=4^{-(x+3)}\cdot\left(3^{x+2}+3^{x+4}\right)=\frac{1}{4^{x+3}}\cdot\left(3^{x+3-1}+3^{x+3+1} \right)$$
Now try to factor out a term with $3^{(\ldots)}$ and see if you can apply some exponent laws.
Hirshy
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