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Given that two to the π₯-power minus two to the π¦-power equals seven and two to the π₯-power plus two to the π¦-power equals nine, find the value of π₯ and the value of π¦.
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What this question is telling us is we want the value of π₯ and π¦ that make both of these statements true at the same time; weβre looking for their solution.
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One way to find their solution would be to combine these two equations.
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When we do that, we get two to the π₯-power plus two to the π₯-power.
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And the simplest way to write that would be two times two to the π₯-power.
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After that, negative two to the π¦-power plus positive two to the π¦-power equals zero, and seven plus nine equals 16.
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Now we want to try and isolate π₯.
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We divide both sides of the equation by two, which tell us that two to the π₯-power is equal to eight.
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Since we have a variable as our exponent, we want to try to rewrite eight so that itβs an exponent with a base two.
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We know that two cubed equals eight.
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If we substitute two cubed in place of eight, we can say two to the π₯-power is equal to two cubed, and therefore π₯ must be equal to three.
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To find our π¦-variable, weβll plug in what we know for π₯.
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And if weβve done this correctly, we can plug that π₯-value into both of these equations and solve for π¦, and the π¦-values will be the same for either equation.
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This means weβll have two cubed minus two to the π¦-power equals seven and two cubed plus two to the π¦-power equals nine.
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For the first equation, that is eight minus two to the π¦-power equals seven.
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To isolate π¦, we subtract eight from both sides, and we see that negative two to the π¦-power equals negative one.
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Multiplying through by negative one, which means two to the π¦-power is equal to one.
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And this means that π¦ has to be equal to zero.
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Since we have a base of two, the only exponential value that will make two to the π¦-power equal to one is zero.
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We can check that this is true by solving our second equation.
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Again, weβll need to subtract eight from both sides, and we end up with two to the π¦-power equals one, which means, based on our exponents rules, that π¦ has to be equal to zero.
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Weβre saying that the values π₯ equals three and π¦ equals zero make both of these statements true.