Can someone help me with this question of finding x as exponent?

Question

The equation is:

$$6^{x+1} - 6^x = 3^{x+4} - 3^x$$

I need to find x. I forgot how to use logarithm laws. Help would be appreciated. Thanks.

Answer 1

Divide both sides by $3^x$, which is positive for all $x\in\mathbb R$:

$$6^{x+1} - 6^x = 3^{x+4} - 3^x$$

$$\iff 3^{x}\cdot 3\cdot 2^{x+1}-3^x\cdot 2^x=81 \cdot 3^x - 3^x$$

$$\iff 3\cdot 2^{x+1}-2^x=81-1$$

$$\iff 6\cdot 2^x-2^x=80\iff 5\cdot 2^x =80$$

$$\iff 2^x =16\iff x=\log_2(16)=4$$

Answer 2

Factor out $3^x$:

$$ 6^{x+1}-6^x = 3^{x+4}-3^x $$

$$ 3^x(3 \cdot 2^{x+1}-2^x) = 3^x(3^4-1) $$

$$ 3 \cdot (2 \cdot 2^x) - 2^x = 80 $$

Take it from there?

Answer 3

Equivalent to $6*2^x - 2^x = 3^4 - 1$
solve for $2^x$,
$2^x =a$

$6*a - a = 81 -1$
$5*a= 80$

$a = 16.$

$2^x = 16$, therefore $x = 4$

Answer 4

This means $$ 6^x(6-1) = 3^x(3^4 - 1) \iff \\ 5 \cdot 6^x = 80 \cdot 3^x \iff \\ 5 \cdot 2^x \cdot 3^x = 80 \cdot 3^x \iff \\ 5 \cdot 2^x = 80 \iff \\ 2^x = 16 \Rightarrow \\ x = 4 $$

Answer 5

Let $$3^x=t, 2^x=u$$ Your original equation becomes $$6ut-ut=3^4t-t\Rightarrow5ut=t(3^4-1)\Rightarrow\\ u=\frac{3^4-1}{5}\Rightarrow 2^x=16\Rightarrow x\ln2=\ln16\Rightarrow\\x\ln2=\ln2^4=4\ln2\Rightarrow\\x=4$$