Exponents 101

Algebra Level 2

90 a = 2 9 0 b = 5 4 5 ( 1 − a − b ) ( 2 − 2 a ) = ? \begin{aligned} {90}^a &=& 2 \\ 90^b &=& 5 \\ \large 45^{ \frac{(1-a-b)}{(2-2a)} } &=& \ ? \end{aligned}


The answer is 3.

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Benedict Dimacutac by Benedict Dimacutac , 21, Philippines

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6 solutions

Pi Han Goh
May 7, 2015

Challenge Master:

1 − a − b 2 − 2 a = 1 2 ( 1 − a 1 − a − b 1 − a ) = 1 2 ( 1 − b 1 − a ) \frac{1-a-b}{2-2a} = \frac12 \left (\frac{1-a}{1-a} - \frac{b}{1-a} \right) = \frac12 \left(1 - \frac{b}{1-a} \right) .

b = log ⁡ 90 5 , a = log ⁡ 90 2 b = \log_{90}5, a = \log_{90} 2

⇒ 1 − b 1 − a = 1 − log ⁡ 90 5 1 − log ⁡ 90 2 = 1 − log ⁡ 90 5 log ⁡ 90 90 − log ⁡ 90 2 \Rightarrow 1 - \frac{b}{1-a} = 1 - \frac{ \log_{90}{5} } {1- \log_{90}{2} } = 1 - \frac{ \log_{90}{5} } {\log_{90} 90- \log_{90}{2} }

= 1 − log ⁡ 90 5 log ⁡ 90 45 = 1 − log ⁡ 45 5 = log ⁡ 45 45 − log ⁡ 45 5 = log ⁡ 45 9 = 2 log ⁡ 45 3 = 1 - \frac{ \log_{90}{5} } {\log_{90} 45 } = 1 - \log_{45} 5= \log_{45} 45 - \log_{45} 5 = \log_{45}9 = 2 \log_{45} 3

So, 45 45 power that fraction equals to 4 5 1 2 ⋅ 2 log ⁡ 45 3 = 4 5 log ⁡ 45 3 = 3 45^{\frac 12 \cdot 2 \log_{45} 3 } = 45^{\log_{45} 3} = \boxed3 .

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@Pi Han Goh , we really liked your comment, and have converted it into a solution. If you subscribe to this solution, you will receive notifications about future comments.

Brilliant Mathematics Staff - 6 years, 1 month ago
Ikkyu San
May 3, 2015

From equation 9 0 a = 2 ⇒ a = log ⁡ 90 2 90^a=2\Rightarrow \color{#D61F06}{a=\log_{90}{2}}

From equation 9 0 b = 5 ⇒ b = log ⁡ 90 5 90^b=5\Rightarrow \color{#3D99F6}{b=\log_{90}{5}}

Thus,

4 5 1 − a − b 2 − 2 a = 4 5 1 − log ⁡ 90 2 − log ⁡ 90 5 2 − 2 log ⁡ 90 2 = 4 5 log ⁡ 90 90 − log ⁡ 90 2 − log ⁡ 90 5 log ⁡ 90 9 0 2 − log ⁡ 90 2 2 = 4 5 log ⁡ 90 90 2 5 log ⁡ 90 8100 4 = 4 5 log ⁡ 90 9 log ⁡ 90 2025 = 4 5 2 log ⁡ 90 3 2 log ⁡ 90 45 = 4 5 log ⁡ 90 3 log ⁡ 90 45 = 4 5 log ⁡ 3 log ⁡ 90 log ⁡ 45 log ⁡ 90 = 4 5 log ⁡ 3 log ⁡ 45 = 4 5 log ⁡ 45 3 = 3 \begin{aligned}45^{\dfrac{1-\color{#D61F06}a-\color{#3D99F6}b}{2-2\color{#D61F06}a}}=&\ 45^{\dfrac{1-\color{#D61F06}{\log_{90}{2}}-\color{#3D99F6}{\log_{90}{5}}}{2-2\color{#D61F06}{\log_{90}{2}}}}\\=&\ 45^{\dfrac{\log_{90}{\color{#20A900}{90}}-\log_{90}{\color{#20A900}2}-\log_{90}{\color{#20A900}5}}{\log_{90}{\color{magenta}{90^2}}-\log_{90}{\color{magenta}{2^2}}}}\\=&\ 45^{\dfrac{\log_{90}{\color{#20A900}{\dfrac{\dfrac{90}{2}}{5}}}}{\log_{90}{\color{magenta}{\dfrac{8100}{4}}}}}\\=&\ 45^{\dfrac{\log_{90}{\color{#20A900}9}}{\log_{90}{\color{magenta}{2025}}}}\\=&\ 45^{\dfrac{\color{#624F41}2\log_{90}{3}}{\color{#624F41}2\log_{90}{45}}}\\=&\ 45^{\dfrac{\color{#69047E}{\log_{90}{3}}}{\color{#302B94}{\log_{90}{45}}}}\\=&\ 45^{\dfrac{\color{#69047E}{\dfrac{\log{3}}{\log{90}}}}{\color{#302B94}{\dfrac{\log{45}}{\log{90}}}}}\\=&\ 45^{\dfrac{\log{\color{#69047E}{3}}}{\log{\color{#302B94}{45}}}}\\=&\ 45^{\log_{\color{#302B94}{45}}{\color{#69047E}3}}=\boxed{3}\end{aligned}

7 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Another fantastic solution! There's not many solutions with colours in it, certainly a good read. Great job!

Despite writing a beautiful standard approach to this problem, can you think of a simpler approach? Hint: break up the fraction 1 − a − b 2 − 2 a \frac{1-a-b}{2-2a} into two parts.

Chew-Seong Cheong
Aug 20, 2015

4 5 1 − a − b 2 − 2 a = ( 90 2 ) 1 − a − b 2 ( 1 − a ) = ( 90 9 0 a ) 1 − a − b 2 ( 1 − a ) = 9 0 ( 1 − a ) ( 1 − a − b ) 2 ( 1 − a ) = 9 0 1 − a − b 2 = ( 90 9 0 a 9 0 b ) 1 2 = ( 90 2 × 5 ) 1 2 = 9 = 3 \begin{aligned} 45^{\frac{1-a-b}{2-2a}} & = \left( \frac{90}{2} \right)^{\frac{1-a-b}{2(1-a)}} = \left( \frac{90}{90^a} \right)^{\frac{1-a-b}{2(1-a)}} = 90^{\frac{(1-a)(1-a-b)}{2(1-a)}} = 90^{\frac{1-a-b}{2}} \\ & = \left( \frac{90}{90^a90^b} \right)^{\frac{1}{2}} = \left( \frac{90}{2\times 5} \right)^{\frac{1}{2}} = \sqrt{9} = \boxed{3} \end{aligned}

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Mohamed Wafik
Aug 20, 2015

9 0 1 − a = 90 2 = 45 90^{1-a} = \frac{90}{2} = 45

4 5 1 1 − a = 90 45^{\frac{1}{1-a}} = 90

4 5 1 − a − b 1 − a = 9 0 1 − a − b = 9 0 1 9 0 a + b 45^{\frac{1-a-b}{1-a}} = 90^{1-a-b} = \frac {90^{1}}{90^{a+b}}

= 90 9 0 a × 9 0 b = 90 2 × 5 = 90 10 = 9 = \frac {90}{90^{a} \times 90^{b}} = \frac{90}{2 \times 5} = \frac{90}{10} = 9

4 5 1 − a − b 2 − 2 a = 9 = 3 45^{\frac{1-a-b}{2-2a}} = \sqrt{9} = \boxed{3}

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Nabil Azmi
May 22, 2015

a= log2/log90..........

b=log5/log90...........

a+b= log10/log90............

1-(a+b)=1-log10/log90= log9/log90............

1-a= 1-log2/log90 =log45/log90.............

(1-a-b)/2(1-a)= log9/log90 *log90/2log45=log9/2log45..........

let (1-a-b)/2(1-a)log45=log x..............

(log9)/2=logx...........

log9 =2log x^2...........

x^2=9 ............

X=3

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Purushottam Gupta
May 22, 2015

whats wrong in here ? can somebody correct this looking for a well explaination !!

0 Helpful 0 Interesting 0 Brilliant 0 Confused

You don't necessarily need to use 10 10 as the base of log ⁡ \log .

Pi Han Goh - 6 years ago

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whats wrong , if i do ? their are no rstrictions on base !!

Purushottam Gupta - 6 years ago

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You're purposely setting 1 − a − b 1-a-b equals to 0 0 , see through your steps again. It's a fallacy to choose a base such that you will force out an unwanted value, try an unknown base instead, like what Ikkyu San did.

Pi Han Goh - 6 years ago

a doubt in your solution: if b/a=log(5)/log(2) how can it be: b/1-a=log(5)/1-log(2)

Bharat Naik - 6 years ago

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umm , guess ratio property ! ??

Purushottam Gupta - 6 years ago

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