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Question Number 132198 by benjo_mathlover last updated on 12/Feb/21
If { ((16^(a+b) = ((√2)/2))),((16^(b+c) = 2)),((16^(a+c) = 2(√2))) :} then c = __
$$\mathrm{If}\:\begin{cases}{\mathrm{16}^{{a}+{b}} \:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\\{\mathrm{16}^{{b}+{c}} \:=\:\mathrm{2}}\\{\mathrm{16}^{{a}+{c}} \:=\:\mathrm{2}\sqrt{\mathrm{2}}}\end{cases} \\ $$$$\:\mathrm{then}\:\mathrm{c}\:=\:\_\_\: \\ $$
Answered by EDWIN88 last updated on 12/Feb/21
eq(1) 16^(2(a+b)) = 2^(−1) ⇒8a+8b=−1 eq(2) 16^(b+c) = 2 ⇒4b+4c = 1 eq(3) 16^(a+c) = 2(√2)=2^(3/2) ⇒4a+4c=(3/2) eliminate eq(1)∧eq(2)⇒ 8a+8b=−1 8c+8b=2 _________ − eq(4) 8a−8c = −3 8a+8c = 3 ___________ − −16c = −6 ⇒ c = (3/8)
$$\mathrm{eq}\left(\mathrm{1}\right)\:\mathrm{16}^{\mathrm{2}\left(\mathrm{a}+\mathrm{b}\right)} \:=\:\mathrm{2}^{−\mathrm{1}} \:\Rightarrow\mathrm{8a}+\mathrm{8b}=−\mathrm{1} \\ $$$$\mathrm{eq}\left(\mathrm{2}\right)\:\mathrm{16}^{\mathrm{b}+\mathrm{c}} \:=\:\mathrm{2}\:\Rightarrow\mathrm{4b}+\mathrm{4c}\:=\:\mathrm{1} \\ $$$$\mathrm{eq}\left(\mathrm{3}\right)\:\mathrm{16}^{\mathrm{a}+\mathrm{c}} \:=\:\mathrm{2}\sqrt{\mathrm{2}}=\mathrm{2}^{\mathrm{3}/\mathrm{2}} \:\Rightarrow\mathrm{4a}+\mathrm{4c}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{eliminate}\:\mathrm{eq}\left(\mathrm{1}\right)\wedge\mathrm{eq}\left(\mathrm{2}\right)\Rightarrow\:\mathrm{8a}+\mathrm{8b}=−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{8c}+\mathrm{8b}=\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\_\_\_\_\_\_\_\_\_\:− \\ $$$$\:\mathrm{eq}\left(\mathrm{4}\right)\:\mathrm{8a}−\mathrm{8c}\:=\:−\mathrm{3}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{8a}+\mathrm{8c}\:=\:\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\_\_\_\_\_\_\_\_\_\_\_\:− \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{16c}\:=\:−\mathrm{6}\:\Rightarrow\:\mathrm{c}\:=\:\frac{\mathrm{3}}{\mathrm{8}} \\ $$
Answered by Rasheed.Sindhi last updated on 14/Feb/21
AnOther way { ((16^(a+b) = ((√2)/2)........(i))),((16^(b+c) = 2..........(ii))),((16^(a+c) = 2(√2)......(iii))) :} (i)/(ii): 16^(a−c) =((√2)/4)................(iv) (iii)/(iv): 16^(2c) =2(√2).(4/( (√2)))=8 2^(8c) =2^3 c=3/8
$${AnOther}\:{way} \\ $$$$\begin{cases}{\mathrm{16}^{{a}+{b}} \:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}……..\left({i}\right)}\\{\mathrm{16}^{{b}+{c}} \:=\:\mathrm{2}……….\left({ii}\right)}\\{\mathrm{16}^{{a}+{c}} \:=\:\mathrm{2}\sqrt{\mathrm{2}}……\left({iii}\right)}\end{cases}\: \\ $$$$\left({i}\right)/\left({ii}\right): \\ $$$$\mathrm{16}^{{a}−{c}} =\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}…………….\left({iv}\right) \\ $$$$\left({iii}\right)/\left({iv}\right): \\ $$$$\mathrm{16}^{\mathrm{2}{c}} =\mathrm{2}\sqrt{\mathrm{2}}.\frac{\mathrm{4}}{\:\sqrt{\mathrm{2}}}=\mathrm{8} \\ $$$$\mathrm{2}^{\mathrm{8}{c}} =\mathrm{2}^{\mathrm{3}} \\ $$$${c}=\mathrm{3}/\mathrm{8} \\ $$$$ \\ $$

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