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

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

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