To solve exponential equations like 5^x - 100 = 0, isolate the exponential term, express the constant as a product involving the base (100 = 5² × 4), take logarithms of both sides, apply logarithm laws (log(ab) = log(a) + log(b) and log(a^b) = b·log(a)), and solve for x using the change of base formula to get x = 2 + log₅(4).
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Olympiad Mathematics | Indian | Can you solve this? | VerifiedAdded:
Hi everyone.
Let me show you how to solve this problem here.
Solution.
The equation is 5 to the power of x - 100 = 0.
Okay, so what do we do first?
Move this to the other side, right?
Remember the additive inverse of -100 is +100.
So, the additive inverse is what will be on the right-hand side, and that is 100.
And 100 + 0 will still give us 100.
Okay, so from here we look at the 100 over here.
Now, can it be in the base of 5?
The answer is no.
But we know that 5 * 20 will give us 5 um 100.
And 5 is still a factor of 20.
So, 5 into 20 will give 4, right?
So, 5 * 5 * 4 is 100.
And at this point we say that 5 to the power of x is equal to 5 squared * by 4.
So, that we will now take um the log of both sides. Since the base on the right is not equal to the base on the left completely.
So, we take we take the log of both sides.
log 5 to power x is equal to log 5 squared * 4.
So, from here now we are going to apply some laws of logarithm.
Especially to the right-hand side.
There are two laws I'm going to apply to the right-hand side.
Now, the first law log a b is equal to log a + log b.
Okay?
And then, that means that log 5 to power x will be equal to log Our a now is 5 squared, so we put 5 squared here.
And our b is 4.
+ log 4.
Now, what do I do? I told you two laws will be applicable to the right-hand side.
And one will be applied to the left.
And we also know that log ab log a to power b is the same as b log a.
So, this law will apply to this and also apply to this.
Because both of them have powers.
The x here is the power, so it goes behind to give x log 5.
Then, two here is the power, so it gives us 2 log 5.
Then, we have our + + log 4, right?
So, we write our log 4.
And um from here we are going to divide both sides of the equation by log 5.
I will let you know why we are dividing by log 5.
So, that this log 5 can cancel this one.
So, only x remains on the left-hand side now.
And our x will be two log 5 + log 4 all over log 5.
Okay?
And then, if we go on from here, we can distribute this um log 5.
So, that it will work for both both equation um both numerators.
So, let's do that.
Okay, so from here now this log 5 will cancel that.
So, that we have x to be equal to 2 + log 4 / log 5.
Now, from here again, we can apply change of base to this.
Because we know that if we have log a / log b both of them are in the same base.
Now, we can change the base and this will be equal to log a to base b.
Okay, so I will apply the same thing to this.
So, that we have x to be two + log 4 to base 5.
Okay, so the 5 becomes the base.
And this right now is our value of x.
We want to put this back into the equation, which is 5 to the power of x - 100 = 0.
So, this means that 5 to the power of x here is expected to be 100.
Because it is 100 - 100 that should be 0.
So, let's come down to 5 to the power of x.
5 to the power of x will now be 5 to the power of 2 + log 4 to base 5.
So, that the um our x is now replaced over here.
And then, if you go on again, we can write this as 5 squared * 5 to the power of log 4.
Both to base 5.
Okay, so this is what we have.
And there is a law that says Let me write it here. There's a law again that says a to the power of log b to base a is equal to b.
Right? This is one of the laws of logarithm.
So, if this is true, then we can relate it to what we have over here.
So, that we have 5 to the power of 2 * by the whole of this now will be 4.
Comparing it to what we have here.
And 5 squared is 25 * which is 100.
So, that means that we are very correct to say the value of x is equal to 2 + log 4 to base 5.
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