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Can anyone help me on this? It is for a 8th grader.

What is the value of $x$ in $222^x-111^x*7=111^x$?

I know the equation can be rearranged as $222^x=111^x*7-111^x=6*111^x$. Then what is next?

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  • $\begingroup$ Check your rearranged answer,it is 8 not 6. $\endgroup$ Sep 22 '16 at 16:59
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Divide through by $111^x$ and you get $$2^x - 7 = 1$$

Rearranging yields: $$2^x = 8$$

So $x=3$

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Actually you got the first step backwards.

$222^x - 7 \cdot 111^x = 111^x$

$222^x = 8 \cdot 111^x$

$(2 \cdot 111)^x = 8 \cdot 111^x$

$2^x \cdot 111^x = 8 \cdot 111^x$

$2^x = 8$

$x = 3$

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$$222^x-111^x\cdot7=111^x$$

$$\implies(2\times111)^x-111^x\cdot7=111^x$$

$$\implies2^x\cdot111^x-7\cdot111^x=111^x$$

$$\implies111^x(2^x-7)=111^x$$

$$\implies2^x-7=1$$

$$2^x=8$$

$$\implies x=3$$

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Set

$$111 = a$$

hence

$$222 = 2a$$

Now your equation is simpler:

$$(2a)^x -7 a^x = a^x$$

hence

$$(2a)^x = 8a^x$$

$$2^xa^x = 8a^x$$

Since the exponential is never zero (and here $a = 111$) divide by $a^x$

$$2^x = 8$$

Hence with the logarithm base $2$:

$$\lg(2^x) = \lg(8)$$

$$x = 3$$

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