Question Video: Finding the Solution Set of Exponential Equations Involving Logarithms over the Set of Real Numbers | Nagwa Question Video: Finding the Solution Set of Exponential Equations Involving Logarithms over the Set of Real Numbers | Nagwa

Question Video: Finding the Solution Set of Exponential Equations Involving Logarithms over the Set of Real Numbers Mathematics • Second Year of Secondary School

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Find the solution set of ๐ฅ^(log_(๐ฅ) ๐ฅโถ) = log 10โถโด in โ.

02:18

Video Transcript

Find the solution set of ๐ฅ to the power of log base ๐ฅ of ๐ฅ to the sixth power equals log of 10 to the power of 64.

Here we have a rather nasty-looking logarithmic equation. Letโs begin by seeing if thereโs any way we can simplify it somewhat. Well, first, we recall one of our laws of logarithms. And this says that log of ๐ to the power of ๐ is the same as ๐ to the power of log ๐. And it doesnโt matter what the base of this log is. This means we can rewrite the right-hand side of our equation as 64 log of 10.

Now, if no base is included, we can assume that the base of our logarithm is 10. So this is actually 64 log base 10 of 10. But we know that log base ๐ of ๐ is simply one. So 64 log base 10 of 10 or 64 log of 10 is 64 times one, which is just 64. And so, weโve simplified the right-hand side of our equation. We get ๐ฅ to the power of log base ๐ฅ of ๐ฅ to the sixth power equals 64. But can we simplify the left-hand side?

Well, weโre actually going to use the same rules. This time, weโre simply going to consider the exponent. Itโs log base ๐ฅ of ๐ฅ to the sixth power. Using our first rule, we see that log base ๐ฅ of ๐ฅ to the sixth power is the same as six log base ๐ฅ of ๐ฅ. Then, using our second rule, we see that log base ๐ฅ of ๐ฅ is one. So six log base ๐ฅ of ๐ฅ is six times one, which is simply six. And so, our equation becomes ๐ฅ to the sixth power equals 64.

Now, to solve for ๐ฅ, we could take the sixth root of both sides, remembering, of course, that we take both the positive and negative sixth root of 64. But actually, we know that two to the sixth power is equal to 64. So ๐ฅ is equal to two. Weโre not actually going to include the solution ๐ฅ is equal to negative two. And thatโs because the base of a logarithm cannot be negative.

And if we go back to our question, ๐ฅ is indeed the base of one of our logs. And so, the solution to this equation is simply ๐ฅ is equal to two. Using set notation, we say that the solution set of ๐ฅ to the power of log base ๐ฅ of ๐ฅ to the sixth power equals log of 10 to the power of 64 is two.

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