Question Number 60620 by naka3546 last updated on 22/May/19
$$\mathrm{4}^{{x}} \:+\:\mathrm{9}^{{x}} \:+\:\mathrm{25}^{{x}} \:\:=\:\:\mathrm{6}^{{x}} \:+\:\mathrm{10}^{{x}} \:+\:\mathrm{15}^{{x}} \\ $$$${x}\:\:=\:\:? \\ $$
Answered by fjdjdcjv last updated on 22/May/19
$${x}=\mathrm{0} \\ $$
Answered by tanmay last updated on 22/May/19
$$\mathrm{2}^{{x}} ={a}\:\:\mathrm{3}^{{x}} ={b}\:\:\:\mathrm{5}^{{x}} ={c} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} ={ab}+{bc}+{ca} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{c}^{\mathrm{2}} −\mathrm{2}{ab}−\mathrm{2}{bc}−\mathrm{2}{ac}=\mathrm{0} \\ $$$$\left({a}−{b}\right)^{\mathrm{2}} +\left({b}−{c}\right)^{\mathrm{2}} +\left({c}−{a}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${so}\:\left({a}−{b}\right)^{\mathrm{2}} =\mathrm{0}\rightarrow{a}={b} \\ $$$${similarly}\:\left({b}−{c}\right)^{\mathrm{2}} =\mathrm{0}\rightarrow{b}={c} \\ $$$${so}\:{a}={b}={c} \\ $$$${now}\:\mathrm{2}^{{x}} =\mathrm{3}^{{x}} \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}\:}\right)^{{x}} =\mathrm{1}=\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{0}} \:\boldsymbol{{so}}\:\boldsymbol{{x}}=\mathrm{0} \\ $$
Answered by mr W last updated on 24/May/19
$${a}^{{x}} +{b}^{{x}} +{c}^{{x}} ={p}^{{x}} +{q}^{{x}} +{r}^{{x}} \\ $$$${or} \\ $$$${a}^{{x}} +{b}^{{x}} +{c}^{{x}} +{d}^{{x}} ={p}^{{x}} +{q}^{{x}} +{r}^{{x}} +{s}^{{x}} \\ $$$$...... \\ $$$$\Rightarrow{x}=\mathrm{0}\:{is}\:\boldsymbol{{one}}\:{solution} \\ $$
Commented by naka3546 last updated on 23/May/19
$${Thank}\:\:{you},\:\:{sir}. \\ $$
Commented by fjdjdcjv last updated on 23/May/19
$${there}\:{are}\:\mathrm{3}\:{solutions} \\ $$$${but}\:{i}\:{think}\:{these}\: \\ $$$$\mathrm{1}+\mathrm{5}^{{x}} +\mathrm{8}^{{x}} +\mathrm{12}^{{x}} =\mathrm{2}^{{x}} +\mathrm{3}^{{x}} +\mathrm{10}^{{x}} +\mathrm{11}^{{x}} \\ $$$${with}\:{x}=\mathrm{1}\:{x}=\mathrm{2}\:{and}\:{x}=\mathrm{3} \\ $$
Commented by mr W last updated on 24/May/19
$${you}\:{are}\:{right}\:{sir},\:{thanks}! \\ $$