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

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

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Find all the possible values of π₯ for which π₯^(logβ π₯) = 64π₯β»ΒΉ.

05:26

### Video Transcript

Find all the possible values of π₯ for which π₯ to the power of log base two of π₯ is equal to 64 times π₯ to the power of negative one.

In this question, weβre asked to determine all of the possible values of π₯ which satisfy a given equation. And to do this, we can notice this equation is quite difficult to solve. For example, the left-hand side of this equation is an exponential function in π₯. If we try to simplify this equation, we can see this is quite difficult. So instead, we want to take logarithms of both sides of the equation. Before we do this, however, we do need to check that both sides of the equation are positive.

Since the equation involves the log base two of π₯ and we recall we can only take the logarithm of positive numbers, we know π₯ must be positive. Then, a positive number raised to any exponent will be positive. So, the left-hand side of this equation is positive. And we can even say the same about the right-hand side of this equation; itβs 64 times π₯ to the power of negative one. Thatβs 64 divided by π₯, and π₯ is a positive number. Therefore, weβll take the log base two of both sides of the equation, giving us log base two of π₯ to the power of log base two of π₯ is equal to log base two of 64 times π₯ to the power of negative one. And itβs worth noting we could choose any positive number not equal to one for the base of our logarithm. However, we choose two because log base two of π₯ already appears in the equation.

We now need to simplify this equation by using our laws of logarithms. First, on the left-hand side of this equation, weβre taking the logarithms of an exponent. So, weβre going to simplify this by using the power rule for logarithms. This tells us for a positive real number π not equal to one, positive real number π₯, and any real number π, the log base π of π₯ to the power of π is equal to π times the log base π of π₯. And we know all of these hold true on the left-hand side of our equation. Our value of π, the base of the logarithm, is two. Our value of π₯ has already been shown to be positive and log base two of π₯ is well defined; itβs a real number. Therefore, by applying the power rule for logarithms to the left-hand side of our equation, we get log base two of π₯ multiplied by log base two of π₯.

Next, on the right-hand side of this equation, we see weβre taking the logarithm of a product, so weβll simplify this by using the productβs rule for logarithms. This tells us for any positive real numbers π₯ one and π₯ two and positive real number π not equal to one, the log base π of π₯ one times π₯ two is equal to the log base π of π₯ one plus the log base π of π₯ two. And once again, we know all of these conditions hold true. The value of π, the base, is two, the value of π₯ one is 64, and the value of π₯ two is π₯ to the power of negative one, where we know π₯ is positive. Therefore, by applying the product rule for logarithms to the right-hand side of our equation, it simplifies to give us log base two of 64 plus log base two of π₯ to the power of negative one. However, we still canβt solve this equation for π₯, so letβs simplify even further.

First, on the right-hand side of this equation, we can see weβre taking the log base two of 64. We can evaluate this by noting two to the sixth power is equal to 64, so log base two of 64 is equal to six. Next, we can simplify the second term on the right-hand side of our equation by once again using the power rule for logarithms, where our value of π is two, π₯ is equal to π₯, and our exponent π is negative one. Therefore, this term simplifies to give us negative log base two of π₯. Therefore, weβve simplified our equation to be log base two of π₯ times log base two of π₯ is equal to six minus log base two of π₯.

And now, we can notice something interesting. Our equation is written entirely in terms of the log base two of π₯. Therefore, we can simplify this equation by substitution. Weβre going to substitute π¦ is equal to the log base two of π₯. The left-hand side of this equation is log base two of π₯ multiplied by log base two of π₯. Thatβs π¦ multiplied by π¦, which is π¦ squared. Similarly, the right-hand side of this equation simplifies to give us six minus π¦. Therefore, we need to find the values of π¦ which satisfy π¦ squared is equal to six minus π¦. We can solve this since this is a quadratic equation. First, we add π¦ to both sides of the equation and subtract six from both sides of the equation, giving us π¦ squared plus π¦ minus six is equal to zero.

Then, we just need to solve this quadratic equation. One way of doing this is by factoring. We notice three times negative two is negative six and three plus negative two is equal to one. Therefore, we can factor this quadratic as π¦ plus three multiplied by π¦ minus two, and this needs to be equal to zero. Finally, for the product of two factors to be equal to zero, one of the two factors must be equal to zero. Therefore, we either have π¦ is negative three or π¦ is equal to two. But remember, weβre asked to find the values of π₯ which solve the equation, not the value of π¦. So, we need to use our substitution. Substituting π¦ is equal to log base two of π₯ into these two solutions gives us two solutions. Log base two of π₯ is equal to negative three or log base two of π₯ is equal to two.

And now, we can solve both of these two equations for π₯. If log base two of π₯ is equal to negative three, then we know π₯ will be equal to two to the power of negative three. Similarly, if log base two of π₯ is equal to two, then π₯ is equal to two squared. And two to the power of negative three is one-eighth and two squared is four. Therefore, either π₯ is equal to four or one-eighth. And we can verify both of these solutions by substituting them back into our equation.

Therefore, we were able to find all of the possible values of π₯ which solve the equation π₯ to the power of log base two of π₯ is equal to 64 times π₯ to the power of negative one. Either π₯ is equal to four or π₯ is equal to one-eighth.

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