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Solve the following equation using Logarithms and leave your answer to 2 significant figures.

$4^{x}-2^{x+1}-3 = 0$

I tried to change the expression into a quadratic expression but I am stuck on where I found $3 = 2^{x}$.

Patiently waiting for your help mates

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  • $\begingroup$ From $2^x=3$ you can just take logarithms on both sides, and use the basic properties of logarithms to simplify $\log(2^x)$. $\endgroup$ Mar 13, 2021 at 10:37
  • $\begingroup$ I am stuck on how the properties of logarithms work ,if you don't mind help me solving the whole problem please 🙏 $\endgroup$ Mar 13, 2021 at 10:43
  • $\begingroup$ 1. What properties of logarithms do you know? 2. How did you get to $2^x=3$? $\endgroup$ Mar 13, 2021 at 10:46

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When you found $2^x=3$, by the very definition, it follows that $$\boxed{x=\log_23}$$


NOTE:

Logarithms are defined as $$a^b=c\iff \log_ac=b$$ Here, $a$ is the base. For $a=10$, we call it common logarithm, simply denoted by $\log x$ ($=\log_{10}x$). For $a=e$, we call it natural logarithm denoted by $\ln x$ ($=\log_ex$).

Since you are a beginner to the concept, I recommend you to check out some easy introduction to them, such as by mathsisfun.com or brilliant.org.


Hope this helps. Ask anything if not clear :)

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  • $\begingroup$ @BiliatLigomeka: I edited my answer. Please check that out! $\endgroup$ Mar 13, 2021 at 11:10
  • $\begingroup$ So here we have log 3 base 2 . how can I find the value of this on a calculator ? I am failing to input the base of the log $\endgroup$ Mar 13, 2021 at 11:19
  • $\begingroup$ So here we have log 3 base 2 . how can I find the value of this on a calculator ? I am failing to input the base of the log $\endgroup$ Mar 13, 2021 at 11:20
  • $\begingroup$ @BiliatLigomeka: Use a scientific calculator like HiPER Scientific Calculator (available on play store), or an online engine like Wolfram Alpha. WA is highly recommended as every student/researcher uses it for scientific computation. $\endgroup$ Mar 13, 2021 at 13:44
  • $\begingroup$ Don't use logs to base two, if your calculator doesn't have them. Instead, from $2^x=3$ you get $\log(2^x)=\log3$, then by properties of logs you get $x\log2=\log3$, so $x=(\log3)/(\log2)$. Now you can use any log function your calculator has to get as many decimals of $x$ as your calculator will give you. $\endgroup$ Mar 13, 2021 at 22:39
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Just take the log base 2 on both sides. You would get $x=log_2 3$.

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  • $\begingroup$ Anyone to get me through the whole question please $\endgroup$ Mar 13, 2021 at 10:45
  • $\begingroup$ It's better if you do some of the work, Biliat, and show us what you've done and what you know and where you get stuck. $\endgroup$ Mar 13, 2021 at 10:47
  • $\begingroup$ Alright,I simplified the problem using the laws of indices such that 4^x turned out to be 2 ^2x . Now keeping in mind that 2^x+1 = 2^x + 2^1 I splitted it. $\endgroup$ Mar 13, 2021 at 10:55
  • $\begingroup$ Sure, you must have gotten a quadratic, in $2^{x}$, and you solved it, getting the value of $2^{x}$ as 3. From then on, you just need to take the logarithms on both sides. If you don't know what they are, this may help, so take a quick look - mathsisfun.com/algebra/logarithms.html. Otherwise, user @ultralegend5385 answers it better than I have. $\endgroup$ Mar 13, 2021 at 11:02
  • $\begingroup$ By 2^x+1 I guess you mean $2^{x+1}$ (you would do well to take the time to learn how to format mathematics on this site – there is help available, through the Help menu). By 2^x+2^1 I hope you mean the product of $2^x$ and $2^1$ (and not the sum, as you have written). Anyway, is your question answered now, or is there still something that needs clarification for you? $\endgroup$ Mar 13, 2021 at 22:35

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