To solve exponential equations with different bases, apply the law of indices (M^(A+B) = M^A × M^B) to separate variables, then divide both sides to isolate terms with the variable, combine bases by subtracting exponents, and take logarithms of both sides to solve for the variable. For the equation 2^(X+3) = 3^(X+1), this method yields X ≈ 2.42.
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Olympiad Mathematics | Russian | Can You Solve This?Añadido:
Hi everyone.
Let's let me show you how to solve this problem very quickly.
Okay, we have two to the power of X plus three equals three to the power of X plus one.
Now, what do we do first?
Do you remember this law of indices?
M to the power of A plus B that this can be written as M to power A times M to power of B.
So, if this is true, then what we have on both sides of the equation will be in this form.
And we will have two to the power of X multiplied by the same two to the power of three.
Now, this is equal to three to the power of X multiplied by the same three to the power of one.
So, what's are we saying?
We are having two to the power of X multiplied by eight because two to power three is eight.
And this is equal to three to the power of X multiplied by three because three to the power of one is the same as three.
Now, I'm thinking what we're going to do next.
Um here is what we are going to do very quickly. We're going to divide both sides of the equation by eight.
Okay, so that these can go.
And we'll be having um two to the power of X to be equal to three to the power of X multiplied by three over eight.
Okay. So, the next thing we're going to do is to remove or bring the terms with X to the same sides of the equation.
So, we don't take these to this side.
Rather, we take these to this side so that three over eight will be alone and this will work with this.
So, we're going to divide this by three to the power of X.
Then divide the left-hand side by three to the power of X. Three to the power of X is canceling each other.
And we'll have two to the power of X.
Okay, two to the power of X over three to the power of X and is equal to three over eight.
Now, we can compare or better still combine what we have here.
The [snorts] numerator has power of X and the denominator has power of X.
So, the combination now will give us two over three to the power of X to be equal to three over eight.
Okay. And now, because the bases are not the same we are expected to take the log of both sides.
So, that means we have the log of two over three to the power of X to be equal to the log of three over eight. Let's close this.
Okay, so let's continue from here.
Okay, so from here the power X will go behind and we'll have log three over Okay, that's supposed to be two over three.
So, we have log two over three.
And that will be equal to log three over eight on the other side, right?
And then there's another law again that we can apply over here.
If we have log C over D you know, this can be written as log C minus log D from one of the laws of logarithm.
So, this means that we are having um X multiplying log two minus log three.
Then on the right-hand side, we have log three minus log eight.
By the way, this is subtraction, right?
Okay, so we continue with this.
We want to find the value of X.
So, there'll be need for us to divide both sides by this.
So, we have log two minus log three.
Then the same thing happens there.
Log two minus log three.
So, that this one can cancel this out.
Okay, so the value of X is log three minus log eight.
Okay, log three minus log eight all over log two minus log three.
Okay, let me show that.
So, this is what it is.
And from our calculator, log three minus log eight will give us negative 0.4 the negative 0.426 approximately.
And then log two minus log three will give negative 0.
Okay, let me confirm that. 0.176 176.
Okay, so this is what we have.
And the division here will give us X to be equal to 2.42.
In fact, approximately 2.42.
So, this is the approximated value of X.
Now, let's put this back into the equation very quickly to see if we're going to have the you know, the value of X mean the value on the left on the left-hand side and that on the right-hand side.
Let's see if the two values can be approximated you know, can be approximately equal to each other.
Okay, so this is the equation.
And our X is 2.4 two, right? So, that means we're having two to the power of 2.42 plus three.
Now, on the right-hand side, we have three to the power of two 2.42 then plus one.
Now, on the left, we have three two to the power of five.42.
That is 2.42 plus three.
Then on the right-hand side, we have three to the power of 3.42.
Now, let's use calculator and see what we are going to have.
On the left-hand side two to the power of 5.2 two to the power of 5.42 is 42.8.
Okay, and this continues, right?
Then on the other hand we're going to have three to the power of four three to the power of 3.42 and that is 48.
48. Okay, sorry, 42.
42.8 as well.
Okay, so this means that our value of X is satisfying.
So, X to be equal to 2.42 satisfies the equation.
Thank you for watching. If you enjoyed yourself subscribe so that I can we can have more videos like this.
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