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$ 9^x-6^x=4^{x+1/2}$, solve for $x$

Please don't solve this problems entirely, I just want some hints. I have tried to substitute $3^x=b$ and $2^x=a$.

Gerry Myerson
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manitsez
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  • Firstly, posting one problem in one question is enough. You should make an edit. The other two problems can be asked in two separate questions. Secondly, show what you have tried. – callculus42 May 09 '16 at 12:25
  • Note: I reformatted your question fairly heavily. Please check to see that I didn't change your meaning. – lulu May 09 '16 at 12:27
  • thanks, but I just edited the question to show only 1 problem. – manitsez May 09 '16 at 12:28
  • Ok that is a good start. You should find $(b-2a)(b+a)=0$. You obviously cannot have $a,b$ of opposite signs, so that gives $b=2a$. – almagest May 09 '16 at 12:32
  • See http://math.stackexchange.com/questions/384090/find-all-real-numbers-x-for-which-frac8x27x12x18x-frac76 – lab bhattacharjee May 09 '16 at 13:25
  • Is that $4^\frac{x+1}{2}$ or $4^{x+\frac{1}{2}}$? –  Jun 16 '21 at 10:19

3 Answers3

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Try dividing $9^x $ throughout, and substitute $(\frac {2}{3} )^x $ as $y $. Also, notice that $\frac {4}{9} = (\frac {2}{3})^2 $.

Shourya Pandey
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  • Alternatively, divide by $6^x$, you will get an equation involving $\left( \frac 2 3 \right)^x$ and $\left( \frac 3 2 \right)^x$. – Alex M. May 09 '16 at 13:02
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Hint: $\ln (y ^x) =x\ln y$. Now you should see the first one.

S.C.B.
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Niklas Rosencrantz
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Consider this: The equation can be reduced to $$9^x - 6^x = 4^x.4^{1/2}$$ which is $$9^x = 6^x + 2.4^x$$ Now the RHS is divisible by $2$, but the LHS is not, for any values of $x\in \mathbb Z^+$.
Rest, try using $ln(y^x) = xln(y)$ as others have suggested. That should give the final answer

Siddd
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