# Finding the value of $x$ in an exponential form

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

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

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 :)

• @BiliatLigomeka: I edited my answer. Please check that out! Mar 13, 2021 at 11:10
• 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 Mar 13, 2021 at 11:19
• 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 Mar 13, 2021 at 11:20
• @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. Mar 13, 2021 at 13:44
• 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. Mar 13, 2021 at 22:39

Just take the log base 2 on both sides. You would get $$x=log_2 3$$.

• Anyone to get me through the whole question please Mar 13, 2021 at 10:45
• 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. Mar 13, 2021 at 10:47
• 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. Mar 13, 2021 at 10:55
• 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. Mar 13, 2021 at 11:02
• 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? Mar 13, 2021 at 22:35