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How to solve this exponential equation?

$$7 \cdot 3^{x+1} - 5^{x+2}= 3^{x+4}- 5^{x+3}$$

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    Good first step is to write everything using only $x$ in the exponents. $7\cdot 3^{x+1}$ for example can be rewritten as $7\cdot 3\cdot 3^x$ and $5^{x+2}$ can be rewritten as $5\cdot 5\cdot 5^x$. Then try to simplify. Logarithms will likely come into play as well near the end. – 2017-02-01

3 Answers 3

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Group them according to the base:

\begin{align} &7\cdot 3^{x+1}-3^{x+4}=5^{x+2}-5^{x+3}\\ \implies&3^{x+1}(7-3^3)=5^{x+2}(1-5)\\ \implies&3^{x+1}\cdot (-20)=5^{x+2}\cdot(-4)\\ \implies&3^{x+1}\cdot 5=5^{x+2}\\ \implies&3^{x+1}=5^{x+1} \end{align}

Generally, you would now use logarithms, although this case is rather obvious.

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$$\\ 7 \cdot 3^{x+1} - 5^{x+2}= 3^{x+4}- 5^{x+3}\\ 125\cdot { 5 }^{ x }-25\cdot { 5 }^{ x }=81\cdot { 3 }^{ x }-21\cdot { 3 }^{ x }\\ 100\cdot { 5 }^{ x }=60\cdot { 3 }^{ x }\\ { \left( \frac { 5 }{ 3 } \right) }^{ x }=\frac { 3 }{ 5 } \\ x=-1$$

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$$7 \cdot 3^{x+1} - 5^{x+2}= 3^{x+4}- 5^{x+3}$$ $$21 \cdot 3^{x} -25\cdot 5^{x}= 81\cdot3^{x}- 125\cdot5^{x}$$ $$100\cdot 5^{x}= 60\cdot3^{x}$$ $$\left(\frac{5}{3}\right)^x=\frac{3}{5}=\left(\frac{5}{3}\right)^{-1}$$ $$x=-1$$