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Solve each equation for x.

(a) $ 2^{x - 5} = 3 $(b) $ \ln x + \ln (x - 1) = 1 $

(a) $2^{x-5}=3 \Leftrightarrow \log _{2} 3=x-5 \Leftrightarrow x=5+\log _{2} 3$\[O r: 2^{x-5}=3 \Leftrightarrow \ln \left(2^{x-5}\right)=\ln 3 \Leftrightarrow(x-5) \ln 2=\ln 3 \Leftrightarrow x-5=\frac{\ln 3}{\ln 2} \Leftrightarrow x=5+\frac{\ln 3}{\ln 2}\](b) $\ln x+\ln (x-1)=\ln (x(x-1))=1 \Leftrightarrow x(x-1)=e^{2} \Leftrightarrow x^{2}-x-e=0 .$ The quadratic formula (with $a=1$$b=-1,$ and $c=-e$ ) gives $x=\frac{1}{2}(1 \pm \sqrt{1+4 e}),$ but we reject the negative root since the natural logarithm is notdefined for $x<0 .$ So $x=\frac{1}{2}(1+\sqrt{1+4 e})$.

03:33

Jeffrey P.

Calculus 1 / AB

Calculus 2 / BC

Calculus 3

Chapter 1

Functions and Models

Section 5

Inverse Functions and Logarithms

Functions

Integration Techniques

Partial Derivatives

Functions of Several Variables

Jp R.

September 11, 2019

I believe for answer b you can have the negative portion of the +/- ...

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Okay, so let's solve this equation and we need to use logarithms and I'm going to use natural log. Although we could use LA based two or log based 10 or any kind of long he want. If I take a natural log of the left and the natural log of the right, I can then use the power property to bring this ex opponent out to the front. So I have the quantity X minus five times the natural log of two equals a natural log of three. Now we can distribute the natural log of to to both terms in parentheses, and we have X natural log to minus five. Natural log two equals natural log three and we're trying to isolate X. So for the next step, we could add five natural log to to both sides. We have X natural Long two equals Natural log three plus five natural log to finally to get X by itself. Let's divide both sides by natural log to so divide the entire expression by natural log to Okay, there's our exact answer, and we could have a different looking exact answer if we used a different kind of lager of them, but it should be equivalent. And if we put that in the calculator, we should get approximately 6.5850 Okay, now, for the second part, when you have an equation with lager, them's you're going to have to combine them into a single algorithm. So remember, we have the product property, which tells us that if you have a log, plus a log equals the log of the product. So this will be the natural log of the product. X Times X minus one equals one. So that's the natural log of X squared plus minus X equals one. Remember Natural Log, Maine's law based E. So if we rearrange this into its exponential form, we have e to the first power equals X squared minus x be to the first power is just eat so we hav e equals X squared minus X. So now we have a quadratic equation on her hands. And to solve this quadratic equation, let's get everything over to one side and use the quadratic formula. So we have X equals the opposite of be over to a so 1/2 plus or minus the square root of B squared one minus four times a time. See so minus four times one times negative. Eat. It's a little bit weird to have e inside your quadratic formula, but we just have to run with it. So this gives us X equals 1/2 plus or minus the square root of one plus four e over to now. Let's take those answers and approximate them. Put them both into the calculator and let's see what we get. Okay, The decimal approximations for these are 2.2229 and negative 1.2229 Well, we can't have a negative inside of logar them, so we're going to have to eliminate that answer and only keep the positive answer. So eliminate the exact dancer with the negative on it as well. So X equals 1/2 plus the square root of one plus four e over to

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