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Solve $$4^{2x}=8(4^x)−15.$$

Learning advanced functions online and having trouble finding how to solve this? I have a few questions similar to this so i'm hoping to learn the steps on solving this so I can apply it to future equations.

vitamin d
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abrapa99
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    Substitute $z=4^x$ and then use the quadratic formula. – Cornman Jun 07 '21 at 22:03
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    $z^2-8z+15=0$? You don't need the quadratic formula for that.... –  Jun 07 '21 at 22:09
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    Alternatively, you can write the quadratic as $$(4^x)^2-8(4^x)-15=0 , ,$$which factorises nicely. The substitution $z=4^x$ just makes things easier on the eyes. – Joe Jun 07 '21 at 22:11
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    @MatthewDali Guesswork with the Vieta formulas is nice and all, but it might not be the first thing I would suggest to a person who has to ask this question in the first place. – Arthur Jun 07 '21 at 22:17
  • Just a simple observation? But did you study the theory or immediately wanted to solve an exponential equation? But are you a high school student? – Sebastiano Jun 07 '21 at 22:33
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    @Sebastiano I am a mature student studying high school courses however the content that is provided for me is very limited/non-existent in most cases and i'm on my own to search for theories/solutions. There is no text book just course work provided. It's very independent and I do not have a teacher to ask questions. – abrapa99 Jun 07 '21 at 23:50
  • @Arthur: In my experience, factorising equations is taught before the quadratic formula. – Joe Jun 08 '21 at 18:41
  • @Joe It may be what they first learn. But it is not the first thing I would recommend they do. Once the quadratic formula is taught, I think you would be hard pressed to find a teacher who generally recommends that students to go back to Vieta guessing when the opportunity arises. The best students, sure. Not the average student. Having a repertoire of different approaches to any given problem, and choosing the most suitable on the fly, is in my experience not something weak-to-average students are expected to do. – Arthur Jun 08 '21 at 19:24
  • @Arthur: You make a very valid point there. – Joe Jun 08 '21 at 19:29
  • My comment was not a reproach at all but simply I was surprised. I have concluded exponential equations, exponential inequalities with the use of logarithms with my high school students. Probably the book you use or your professor did not provide you with the basic elements to solve logarithmic equations of your question. I always wish you the best. – Sebastiano Jun 08 '21 at 20:11
  • I’m voting to close this question because some of the comments offer generous hints for the OP to continue on his/her own. – Mittens Jun 09 '21 at 17:52

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Equations such as $$ 4^{2x}=8(4^x)-15 $$ are known as 'disguised quadratics'. To see what I mean, add $15-8(4^x)$ to both sides: $$ 4^{2x}-8(4^x)+15=0 $$ and use the index law $(a^b)^c=a^{bc}$ to write the equation as $$ (4^x)^2-8(4^x)+15=0 \, . $$ This is a actually a quadratic equation because there is a squared term, $(4^x)^2$, a linear term, $-8(4^x)$, and a constant $15$. To make things blatantly obvious, make the substitution $z=4^x$ as Cornman has suggested. The equation becomes $$ z^2-8z+15=0 \, . $$ From here you can use any one of the standard methods for solving quadratics. I noticed that the numbers $-3$ and $-5$ sum to $-8$, and multiply to make $15$. This allows the equation to be factorised nicely: $$ (z-3)(z-5)=0 \, . $$ If two different numbers multiply to make $0$, then one of them must be $0$. In this case, $z-3=0$ or $z-5=0$. Hence, $z=3$ or $z=5$. We still haven't finished solving the equation. Since $z=4^x$, either $4^x=3$ or $4^x=5$. To solve $$ 4^x=3 \, , $$ apply the inverse function $\log_4$ to both sides: \begin{align} \log_4(4^x)&=\log_4(3) \\ x &= \log_4(3) \end{align} The same reasoning tells us that the solution of $4^x=5$ is $x=\log_4(5)$. This gives us the solutions $$\boxed{ \;\\[4pt] \quad x=\log_4(3) \text{ or } x=\log_4(5) \, .\quad \\ } $$ Plugging these solutions back into the original equation helps us ensure that they are all valid.

Joe
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