log question (1 Viewer)

Green Yoda · Jul 4, 2015

Question 3D

InteGrand · Jul 4, 2015

Rathin said:
Question 3D
View attachment 32163

$3. (d) $8^{1-x}=16\sqrt{2}$$

$\therefore \log_2 8^{1-x}=\log_2 \left(16\sqrt{2}\right)$

$\Rightarrow (1-x)\log_2 8 = \log_2 16 + \log_2 \sqrt{2}$

$\Rightarrow 3(1-x)=4+\frac{1}{2}$, since $\log_2 8 = 3$, $\log_2 16 = 4$ and $\log_2 \sqrt{2} = \frac{1}{2}$, because $8=2^3,16=2^4$ and $\sqrt{2}=2^{\frac{1}{2}}$

$\Rightarrow 3(1-x)=\frac{9}{2}$

$\Rightarrow 1-x = \frac{3}{2}$

$\Rightarrow x = -\frac{1}{2}.$

InteGrand · Jul 4, 2015

Wait, they wanted you to do it like this:

$8^{1-x}=16\sqrt{2}\Rightarrow 2^{3(1-x)}=2^4\times 2^{\frac{1}{2}}=2^{4 + \frac{1}{2}}$

$\Rightarrow 3(1-x)=4+\frac{1}{2}\Rightarrow x = -\frac{1}{2}.$

InteGrand · Jul 4, 2015

(I.e. express both sides as powers of 2 using index laws and then equate indices.)

Green Yoda · Jul 4, 2015

thank you

Green Yoda · Jul 4, 2015

btw if this question was in an exponential base 10 form and there was a surd 16√2 you would just directly divide log 16√2 / log 8 instead of seperating the 16 and √2 into 2 logs?

InteGrand · Jul 4, 2015

Rathin said:
btw if this question was in an exponential base 10 form and there was a surd 16√2 you would just directly divide log 16√2 / log 8 instead of seperating the 16 and √2 into 2 logs?

The surds they give you would generally be ones that are nice powers of the base. Like 16√2 is a nice power of 2, but not of 10, so they wouldn't give you something like 16√2 for powers of 10 and expect a nice answer.

The general way to solve an equation of the form $a^x=b$ is take log to the base a of both sides. Then use change of base formula to get it into base 10 or e so you can use your calculator to get a numerical approximation.

log question (1 Viewer)

Green Yoda

Hi Φ

InteGrand

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InteGrand

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InteGrand

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Green Yoda

Hi Φ

Green Yoda

Hi Φ

InteGrand

Well-Known Member

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