Problem

Solve each equation for x. (a) $ \ln (x^2 - 1)…

01:59

Problem 51 Hard Difficulty

Solve each equation for x.

(a) $ e^{7 - 4x} = 6 $
(b) $ \ln (3x - 10) = 2 $

Answer

(a) $e^{7-4 x}=6 \Leftrightarrow 7-4 x=\ln 6 \Leftrightarrow 7-\ln 6=4 x \quad \Leftrightarrow \quad x=\frac{1}{4}(7-\ln 6)$
(b) $\ln (3 x-10)=2 \quad \Leftrightarrow \quad 3 x-10=e^{2} \quad \Leftrightarrow \quad 3 x=e^{2}+10 \Leftrightarrow \quad x=1\left(e^{2}+10\right)$

Johns Hopkins University

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Video Transcript

all right. Solving an exponential equation usually involves lager of them's. And since we're using base E here, we're going to involve the natural log. So we take the natural log of both sides. The natural log of the left side equals the natural log of the right side. And on the left, the natural log function is going to cancel with the E to the X function, leaving you with seven minus four. X equals the natural. August 6. We're trying to isolate X, so the next thing we could do is subtract seven from both sides, so negative for X equals the natural log of six minus seven. Now we can divide both sides by four, and we get X equals the natural log of six minus seven, divided by negative four. Now there's nothing wrong with that answer. It's perfectly fine, however, if you want to make it look a little bit simpler, you can multiply the numerator by negative one and the denominator by negative one, and you end up with X equal seven minus natural logs six. Divided by four. That's the exact answer. If you're looking for an approximate answer, you can put that into your calculator and to four decimal places. You get approximately 1.30 to 1. Okay, Now on to part B, we have a lager rhythmic equation, and we end up solving those using exponents. Specifically, we change it to exponential form. So remember, when you see natural log, it means log Basie. Now we're going to rearrange this into its exponential form. The bases e the powers to and the result would be three X minus 10. So let's keep going and isolate X. So we're going to add Tend to both sides and we have e squared plus 10 equals three X and then divide both sides by three and X will be e squared, plus 10/3. That's the exact answer. And if you're looking for an approximate answer, you can put that in your calculator and 24 decimal places. We get 5.7964

Heather Z.

Oregon State University

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Problem

Solve each equation for x. (a) $ \ln (x^2 - 1)…

Problem 51 Hard Difficulty

Solve each equation for x.

(a) $ e^{7 - 4x} = 6 $
(b) $ \ln (3x - 10) = 2 $

Answer

(a) $e^{7-4 x}=6 \Leftrightarrow 7-4 x=\ln 6 \Leftrightarrow 7-\ln 6=4 x \quad \Leftrightarrow \quad x=\frac{1}{4}(7-\ln 6)$
(b) $\ln (3 x-10)=2 \quad \Leftrightarrow \quad 3 x-10=e^{2} \quad \Leftrightarrow \quad 3 x=e^{2}+10 \Leftrightarrow \quad x=1\left(e^{2}+10\right)$

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Vectors Intro

Vector Basics - Example 1

(a) $e^{7-4 x}=6 \Leftrightarrow 7-4 x=\ln 6 \Leftrightarrow 7-\ln 6=4 x \quad \Leftrightarrow \quad x=\frac{1}{4}(7-\ln 6)$(b) $\ln (3 x-10)=2 \quad \Leftrightarrow \quad 3 x-10=e^{2} \quad \Leftrightarrow \quad 3 x=e^{2}+10 \Leftrightarrow \quad x=1\left(e^{2}+10\right)$

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Lily A.

Kayleah T.

Samuel H.

Michael J.

Vectors Intro

Vector Basics - Example 1

Lily A.

Kayleah T.

Samuel H.

Michael J.

(a) $e^{7-4 x}=6 \Leftrightarrow 7-4 x=\ln 6 \Leftrightarrow 7-\ln 6=4 x \quad \Leftrightarrow \quad x=\frac{1}{4}(7-\ln 6)$
(b) $\ln (3 x-10)=2 \quad \Leftrightarrow \quad 3 x-10=e^{2} \quad \Leftrightarrow \quad 3 x=e^{2}+10 \Leftrightarrow \quad x=1\left(e^{2}+10\right)$