Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 132198 by benjo_mathlover last updated on 12/Feb/21

$$\mathrm{If}\{\begin{array}{l}{16}^{a+b}=\frac{\sqrt{2}}{2}\\ {16}^{b+c}=2\\ {16}^{a+c}=2\sqrt{2}\end{array}$$$$\mathrm{then}\mathrm{c}=\mathrm{\_}\mathrm{\_}$$

Answered by EDWIN88 last updated on 12/Feb/21

$$\mathrm{eq}\left(1\right){16}^{2(\mathrm{a}+\mathrm{b})}=2^{-1}\Rightarrow 8\mathrm{a}+8\mathrm{b}=-1$$$$\mathrm{eq}\left(2\right){16}^{\mathrm{b}+\mathrm{c}}=2\Rightarrow 4\mathrm{b}+4\mathrm{c}=1$$$$\mathrm{eq}\left(3\right){16}^{\mathrm{a}+\mathrm{c}}=2\sqrt{2}=2^{3/2}\Rightarrow 4\mathrm{a}+4\mathrm{c}=\frac{3}{2}$$$$\mathrm{eliminate}\mathrm{eq}\left(1\right)\wedge \mathrm{eq}\left(2\right)\Rightarrow 8\mathrm{a}+8\mathrm{b}=-1$$$$8\mathrm{c}+8\mathrm{b}=2$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}-$$$$\mathrm{eq}\left(4\right)8\mathrm{a}-8\mathrm{c}=-3$$$$8\mathrm{a}+8\mathrm{c}=3$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}-$$$$-16\mathrm{c}=-6\Rightarrow \mathrm{c}=\frac{3}{8}$$

Answered by Rasheed.Sindhi last updated on 14/Feb/21

$$AnOtherway$$$$\{\begin{array}{l}{16}^{a+b}=\frac{\sqrt{2}}{2}........\left(i\right)\\ {16}^{b+c}=2..........\left(ii\right)\\ {16}^{a+c}=2\sqrt{2}......\left(iii\right)\end{array}$$$$\left(i\right)/\left(ii\right):$$$${16}^{a-c}=\frac{\sqrt{2}}{4}................\left(iv\right)$$$$\left(iii\right)/\left(iv\right):$$$${16}^{2c}=2\sqrt{2}.\frac{4}{\sqrt{2}}=8$$$$2^{8c}=2^3$$$$c=3/8$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com