How do I solve this question? $$25^{\left(2x-x^2+1\right)}+9^{\left(2x-x^2+1\right)}=34\left(15^{2x-x^2}\right)$$ I tried splitting 15 to 5 and 3 and writing 25 and 9 as the squares of 5 and 3. I got stuck after that, could anyone please help me?
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I see one obvious solution. – TickaJules Mar 16 '22 at 12:56
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Could you please guide me? Like, at least just type out what I should do? – Anonymousstriker38596 Mar 16 '22 at 13:05
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2Try $x=0$. ${}{}$ – Shaun Mar 16 '22 at 13:13
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In the exponent of the rhs, add and subract $1$ – Claude Leibovici Mar 16 '22 at 13:15
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Oh, yeah got it. Thanks. @Claude Leibovici – Anonymousstriker38596 Mar 16 '22 at 13:17
3 Answers
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Hint:
Let $5^{2x-x^2}=a,3^{2x-x^2}=b$
$$0=25a^2-34ab+9b^2=25a(a-b)-9b(a-b)=?$$
Case$\#1$: If $25a=9b,$
$\iff\dfrac{25}9=\dfrac ba$
$\iff\left(\dfrac35\right)^{-2}=\left(\dfrac35\right)^{2x-x^2}\iff\left(\dfrac35\right)^{x^2-2x-2}=1$
Use Find all real numbers $x$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac76$
Case$\#2$: What if $a=b?$
lab bhattacharjee
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Your way of splitting works.
Set $u=x(2-x)\leq 1$ and divide by $15^u = 3^u5^u$. So, you get
$$25t + 9\frac 1t = 34 \text{ with } t= \left(\frac 53\right)^u$$
Solving for $t>0$ gives
$$t= 1 \Rightarrow u=0 \Rightarrow x=0, x=2$$
and
$$t= \frac 9{25} \Rightarrow u=-2 \Rightarrow x=1+\sqrt 3, x=1-\sqrt 3$$
trancelocation
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As @TickaJules pointed out there is one obvious solution. let $\alpha = 2x-x^2$ and rewrite the equation $$ 25*25^{\alpha}+ 9*9^{\alpha}=34*15^{\alpha} $$ Since $25 + 9=34$ then clearly $\alpha =0$ solves the equation with $x=2$
F. Wright
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