Only Smart Students Can Solve This in 10 Seconds!

Added: 2026-05-14

[00:00:01]Let's solve for X in this nice exponential equation.

[00:00:06]We have 4 raised to power X plus 6 raised to power X equals to 9 raised to power X. Solution here.

[00:00:13]First step, we can divide through by 4 raised to power X. We have 4 raised to power X plus 6 raised to power X equals to 9 raised to power X.

[00:00:24]That is, divide this by 4 raised to power X.

[00:00:27]Divide this by 4 raised to power X.

[00:00:30]Also, divide this by 4 raised to power X.

[00:00:34]Then, yeah, 4 raised to power X cancels each other. You have 1.

[00:00:39]Then plus This follows when we have A raised to power N over B raised to power N can write it as A over B all raised to power N.

[00:00:51]Which implies here we have 6 over 4 all raised to power X equals to also here we have 9 over 4 all raised to power X.

[00:01:05]The next step here, we can reduce this and we have 1 plus 2 goes in 6, that's 3. 2 goes in 4, you have 2.

[00:01:14]3 over 2 all raised to power X equals to Yeah, 9 can be written as 3 squared over 4 also 3 4 2 squared which is all raised to power X.

[00:01:29]Then we have With this, these two we can factor it out and this becomes 1 plus 3 over 2 all raised to power X equals to Also here you have 3 over 2 all raised to power 2 which is also raised to power X.

[00:01:51]That is, when we apply the law of indices, A raised to power M raised to power N same thing as A raised to power MN.

[00:02:00]Which also same thing as a raised to the power m all raised to the power m.

[00:02:04]Which means we can interchange this power.

[00:02:08]Then we bring x x inside two house. Now we have one plus 3 over 2 all raised to the power x equals to 3 over 2 all raised to the power x and all raised to the power 2.

[00:02:25]That's this.

[00:02:28]Here we can represent 3 over 2 raised to the power x as a letter. So let's u be equals to 3 over 2 all raised to the power x.

[00:02:39]And this equation becomes 1 plus u equals to u squared.

[00:02:46]Then we can rearrange. Take everything to one side. We have u squared.

[00:02:51]Then minus u now, minus 1 equals to 0.

[00:02:56]That's this.

[00:02:58]Here we have a quadratic equation. We have a equals to 1, b equals to minus 1, and c equals to minus 1.

[00:03:08]Apply the quadratic formula. Here we have u equals to minus b plus or minus square root of b squared minus 4 a c all over 2 a.

[00:03:21]Then here this becomes u equals to minus minus 1 plus or minus square root of minus 1 squared minus 4 times 1 times minus 1 all over 2 times 1.

[00:03:41]Which implies we have u equals to minus times minus, that's plus. We have 1 plus or minus square root of minus 1 squared, that's 1.

[00:03:53]Minus times minus, that's plus.

[00:03:56]4 times 1 times 1, that's still 4, all over 2.

[00:04:01]Which implies we have U equals to 1 plus or minus root 5 over 2.

[00:04:11]Here we have two possible values of U.

[00:04:13]First one, U equals to 1 plus 5 root 5 over 2.

[00:04:21]Or we have U equals to 1 minus root 5 over 2.

[00:04:29]Then since U is given represented as 3 over 2 all raised to power X.

[00:04:37]Of course, this is greater than 0 here.

[00:04:42]And also, what we have for U here is greater than 0.

[00:04:46]So, there will be a real solution here.

[00:04:48]Our U here is less than 0.

[00:04:51]So, we don't have a real solution here.

[00:04:54]So, no real solution on this side.

[00:05:01]Then what we have here, we equate this and we say 3 over 2 all raised to power X equals to 1 plus root 5 all over 2.

[00:05:16]That is we take the log on both sides. Here we have log 3 over 2 all raised to power X equals to log 1 plus root 5 all over 2.

[00:05:34]That is, this follows when we have log M raised to power P which is same thing as P log M.

[00:05:43]This here we have X log 3 over 2 equals to log 1 + root 5 over 2 Then this follows the law of logarithm.

[00:06:01]When we have log A over B, I write this as log A minus log B.

[00:06:11]This here we have X log 3 minus log 2 equals to Here also we have log 1 + root 5 minus log 2 Then we divide both side by log 3 minus log 2. That is divide this side by log 3 minus log 2 Also divide this side by log 3 minus log 2 Which implies this cancelled each other here and we have X equals to log 1 + 5 minus log 2 all over log 3 minus log 2 Of course, we can get the approximate solution here by getting the value of this also and divide to get the approximate solution. So here we can stop here and we have the value of X in this problem.

[00:07:30]So that's from here we see that the value of X for this given problem the same thing as X equals to log 1 + root 5 minus log 2 all over log 3 minus log 2 And thank you for watching.

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