$$9^x-2^{x+\frac{1}{2}}=2^{x+\frac{7}{2}}-3^{2x-1}.$$

The equation states solve for $x$.

What I first did was put like bases together.

$$3^{2x}+3^{2x-1}= 2^{x+\frac{7}{2}}+ 2^{x+\frac{1}{2}}.$$

Then I factored $3^{2x}$ and $2^x$

$$3^{2x}(1+\frac{1}{3})=2^x(2^{\frac{7}{2}}+2^{\frac{1}{2}}),$$

then I got

$$\frac{3^{2x}}{2^x}=9\sqrt{2}.$$

From here I took $\log$s, but the answer wasn't nice.

What to do?

Your last line is wrong.

It should be $$3^{2x-3}=(\sqrt2)^{2x-3},$$ which gives $x=1.5$.

You got: $$3^{2x}\left(1+\frac{1}{3}\right)=2^x(2^{\frac{7}{2}}+2^{\frac{1}{2}})$$ or $$3^{2x-1}=2^{x-2}2^{\frac{1}{2}}(1+8)$$ or $$3^{2x-3}=2^{x-\frac{3}{2}}$$

  • Can you show me a little bit more work? I don't see how you got those exponents sorry. – Savvas Nicolaou Oct 14 at 15:40
  • @Savvas Nicolaou The mistake in the last line only. – Michael Rozenberg Oct 14 at 15:42
  • I think I fixed the mistake? But I still don't see how you got $2x-3$ – Savvas Nicolaou Oct 14 at 15:44
  • @Savvas Nicolaou I added something. See now. – Michael Rozenberg Oct 14 at 15:46
  • Thank you very much. – Savvas Nicolaou Oct 14 at 15:47

If you let $x=u+{1\over2}$, you can get rid of the pesky square roots: the expression simplifies to

$$9^{u+1/2}-2^{u+1}=2^{u+4}-3^{2u}$$

or

$$3\cdot9^u-2\cdot2^u=16\cdot2^u-9^u$$

This simplifies first to $4\cdot9^u=18\cdot2^u$ and then to $9^{u-1}=2^{u-1}$, which clearly implies $u=1$, i.e., $x=3/2$.

$$9^x-2^{x+\frac12}=2^{x+\frac72}-3^{2x-1}$$ $$\to\frac43(9^x)=9(2^{x+\frac12})$$

Hence we form the two iterates:

$$x_{n+1}=\log_9{\bigg[\frac{27}{4}(2^{x_n+\frac12})\bigg]}$$ and $$x_{n+1}=\log_2{\bigg[\frac{4}{27}(9^{x_n})\bigg]}$$

The first gives the solution $x=\frac32$, the second diverges to $-\infty$, and thus there is no second solution

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