1. ## logarithm problems

evaluate
1.1.5*4^(log (base 4*sqrt(2))(3-sqrt(6))-6log(base 8)(sqrt(3)-sqrt(2))

2.5^(log(base(1/5)(1/2))+log(base sqrt(2))(4/(sqrt(7)+sqrt(3))+log(base 1/2)(1/(10+2sqrt(21))

2. Hello, prasum!

It would help if you learned to use the Space Bar.
I must guess what you meant . . .

$\text{2. Evaluate: }\;5^{\log_{\frac{1}{5}}\left(\frac{1}{2}\right)}+ \log_{\sqrt{2}}\!\left(\dfrac{4}{\sqrt{7}-\sqrt{3}}\right) + \log_{\frac{1}{2}}\!\left(\dfrac{1}{10+2\sqrt{21}} \right)$

Each of the three terms takes an enormous amount of simplifying!

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The exponent of the first term: . $\log_{\frac{1}{5}}\!\!\left(\frac{1}{2}\right) \,=\,P$

. . $\left(\frac{1}{5}\right)^P \:=\:\frac{1}{2}\quad\Rightarrow\quad \left(5^{-1}\right)^P \:=\:2^{-1} \quad\Rightarrow\quad 5^{-P} \:=\:2^{-1}$

Take logs (base 5): . $\log_5\!\left(5^{\text{-}P}\right) \:=\:\log_5\!\left(2^{\text{-}1}\right) \quad\Rightarrow\quad \text{-}P\log_5(5) \:=\:\text{-}\log_5(2)$

. . Hence: . $P \:=\:\log_5(2) \quad\Rightarrow\quad \log_{\frac{1}{5}}\left(\frac{1}{2}\right) \:=\:\log_5(2)$

The first term is: . $5^{\log_5(2)} \;=\;2$

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In the second term, rationalize the denominator.

. . $\dfrac{4}{\sqrt{7}-\sqrt{3}} \cdot \dfrac{\sqrt{7}+\sqrt{3}}{\sqrt{7}+\sqrt{3}} \;=\;\dfrac{4(\sqrt{7} + \sqrt{3})}{7-3} \:=\:\sqrt{7} + \sqrt{3}$

We have: . $\log_{\sqrt{2}}(\sqrt{7}+\sqrt{3}) \:=\:Q

. . $(2^{\frac{1}{2}})^Q \:=\:\sqrt{7} + \sqrt{3} \quad\Rightarrow\quad 2^{\frac{Q}{2}} \:=\:\sqrt{7} + \sqrt{3}$

Take logs (base 2): . $\log_2\!\left(2^{\frac{Q}{2}}\right) \:=\:\log_2(\sqrt{7}+\sqrt{3})$

. . $\frac{Q}{2}\log_2(2) \:=\:\log_2(\sqrt{7}+\sqrt{3}) \quad\Rightarrow\quad Q \:=\:2\log_2(\sqrt{7}+\sqrt{3})$

. . $Q \:=\:\log_2(\sqrt{7} + \sqrt{3})^2$

The second term is: . $2\log_2(\sqrt{7} + \sqrt{3})$

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The thrd term is: . $\log_{\frac{1}{2}}\!\left(\dfrac{1}{10+2\sqrt{21}} \right) \:=\:\log_{\frac{1}{2}}\!\dfrac{1}{(\sqrt{7}+\sqrt {3})^2}$

We have: . $\log_{\frac{1}{2}}(\sqrt{7}+\sqrt{3})^{-2} \:=\: R \quad\Rightarrow\quad \left(\frac{1}{2}\right)^R \:=\:(\sqrt{7}+\sqrt{3})^{-2}$

. . $\left(2^{-1}\right)^R \:=\:(\sqrt{7}+\sqrt{3})^{-2} \quad\Rightarrow\quad 2^{-R} \:=\:(\sqrt{7}+\sqrt{3})^{-2}$

Take logs (base 2): . $\log_2(2^{-R}) \:=\:\log_2(\sqrt{7}+\sqrt{3})^{-2}$

. . $-R\log_2(2) \:=\:-2\log_2(\sqrt{7}+\sqrt{3})$

. . . . . . . . $R \;=\;\2\log_2(\sqrt{7}+\sqrt{3})$

The third term is: . $2\log_2(\sqrt{7}+\sqrt{3})$

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The original expression is: . $2 + 2\log_2\!\left(\sqrt{7}+\sqrt{3}\right) + 2\log_2\!\left(\sqrt{7}+\sqrt{3}\right)$

. . $=\;\log_2(4) + 4\log_2\!\left(\sqrt{7}+\sqrt{3}\right)$

. . $=\;\log_2\!\bigg[4\left(\sqrt{7}+\sqrt{3}\right)^4\bigg]$

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But check my work . . . please!

3. Originally Posted by Soroban
Hello, prasum!

It would help if you learned to use the Space Bar.
I must guess what you meant . . .

Each of the three terms takes an enormous amount of simplifying!

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The exponent of the first term: . $\log_{\frac{1}{5}}\!\!\left(\frac{1}{2}\right) \,=\,P$

. . $\left(\frac{1}{5}\right)^P \:=\:\frac{1}{2}\quad\Rightarrow\quad \left(5^{-1}\right)^P \:=\:2^{-1} \quad\Rightarrow\quad 5^{-P} \:=\:2^{-1}$

Take logs (base 5): . $\log_5\!\left(5^{\text{-}P}\right) \:=\:\log_5\!\left(2^{\text{-}1}\right) \quad\Rightarrow\quad \text{-}P\log_5(5) \:=\:\text{-}\log_5(2)$

. . Hence: . $P \:=\:\log_5(2) \quad\Rightarrow\quad \log_{\frac{1}{5}}\left(\frac{1}{2}\right) \:=\:\log_5(2)$

The first term is: . $5^{\log_5(2)} \;=\;2$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

In the second term, rationalize the denominator.

. . $\dfrac{4}{\sqrt{7}-\sqrt{3}} \cdot \dfrac{\sqrt{7}+\sqrt{3}}{\sqrt{7}+\sqrt{3}} \;=\;\dfrac{4(\sqrt{7} + \sqrt{3})}{7-3} \:=\:\sqrt{7} + \sqrt{3}$

We have: . $\log_{\sqrt{2}}(\sqrt{7}+\sqrt{3}) \:=\:Q

. . $(2^{\frac{1}{2}})^Q \:=\:\sqrt{7} + \sqrt{3} \quad\Rightarrow\quad 2^{\frac{Q}{2}} \:=\:\sqrt{7} + \sqrt{3}$

Take logs (base 2): . $\log_2\!\left(2^{\frac{Q}{2}}\right) \:=\:\log_2(\sqrt{7}+\sqrt{3})$

. . $\frac{Q}{2}\log_2(2) \:=\:\log_2(\sqrt{7}+\sqrt{3}) \quad\Rightarrow\quad Q \:=\:2\log_2(\sqrt{7}+\sqrt{3})$

. . $Q \:=\:\log_2(\sqrt{7} + \sqrt{3})^2$

The second term is: . $2\log_2(\sqrt{7} + \sqrt{3})$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The thrd term is: . $\log_{\frac{1}{2}}\!\left(\dfrac{1}{10+2\sqrt{21}} \right) \:=\:\log_{\frac{1}{2}}\!\dfrac{1}{(\sqrt{7}+\sqrt {3})^2}$

We have: . $\log_{\frac{1}{2}}(\sqrt{7}+\sqrt{3})^{-2} \:=\: R \quad\Rightarrow\quad \left(\frac{1}{2}\right)^R \:=\\sqrt{7}+\sqrt{3})^{-2} " alt="\log_{\frac{1}{2}}(\sqrt{7}+\sqrt{3})^{-2} \:=\: R \quad\Rightarrow\quad \left(\frac{1}{2}\right)^R \:=\\sqrt{7}+\sqrt{3})^{-2} " />

. . $\left(2^{-1}\right)^R \:=\\sqrt{7}+\sqrt{3})^{-2} \quad\Rightarrow\quad 2^{-R} \:=\\sqrt{7}+\sqrt{3})^{-2} " alt="\left(2^{-1}\right)^R \:=\\sqrt{7}+\sqrt{3})^{-2} \quad\Rightarrow\quad 2^{-R} \:=\\sqrt{7}+\sqrt{3})^{-2} " />

Take logs (base 2): . $\log_2(2^{-R}) \:=\:\log_2(\sqrt{7}+\sqrt{3})^{-2}$

. . $-R\log_2(2) \:=\:-2\log_2(\sqrt{7}+\sqrt{3})$

. . . . . . . . $R \;=\;\2\log_2(\sqrt{7}+\sqrt{3})$

The third term is: . $2\log_2(\sqrt{7}+\sqrt{3})$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The original expression is: . $2 + 2\log_2\!\left(\sqrt{7}+\sqrt{3}\right) + 2\log_2\!\left(\sqrt{7}+\sqrt{3}\right)$

. . $=\;\log_2(4) + 4\log_2\!\left(\sqrt{7}+\sqrt{3}\right)$

. . $=\;\log_2\!\bigg[4\left(\sqrt{7}+\sqrt{3}\right)^4\bigg]$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

But check my work . . . please!
thanks