Question

Solve ;)
${7}^{1 + x} + {7}^{1 - x} = 50$
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Answer 1

Answer:

$x=-1\;or\;1$

Step-by-step explanation:

Given :-

$7^1^+^x+7^1^-^x\;=\;50$

To find :-

$x$

Solution :-

We can write $7^1^+^x+7^1^-^x\;=\;50$ as,

$7.7^x+\frac{7}{7^x}-50\;=\;0$

That is,

$7.7^x7^x+7-50.7^x\;=\;0$

$7.7^2^x-50.7^x+7\;=\;0$

Take $7^x\;=\;t$

$7t^2-50t+7\;=\;0$

$7t^2-49t-t+7\;=\;0$

$7t(t-7)-1(t-7)\;=\;0$

$(7t-1)(t-7)\;=\;0$

∴ $t=\frac{1}{7}\;or\;t=7$ are the roots of the above equation.

So, we get,

$7^x=\frac{1}{7}\;or\;7^x=7$

$7^x=7^-^1\;or\;7^x=7^1$

∴ $x=-1\;or\;1$