eksponential funktion

**skaters** · September 22nd 2012, 09:24 AM

Im stuck with some examples - I would be happy to get some help/guidance

1)(0,16)^x = 5/2
2) 2^x+3*2^x-3 =22
3) 5*0,2^2-x = (1/25)^2x-1
4)(1/3)^2-x -6*9^x-4/2 +2*3^x-6=29

**Soroban** · September 22nd 2012, 12:24 PM

Hello, skaters!

These are not easy ones . . . Here is some help.

$1)\;(0.16)^x \:=\:\tfrac{5}{2}$

The left side is: . $(0.16)^x \:=\:\left(\tfrac{16}{100}\right)^x \;=\;\left(\tfrac{4}{25}\right)^x \;=\; \left[\left(\tfrac{2}{5}\right)^2\right]^x \;=\;\left(\tfrac{2}{5}\right)^{2x}$

The right side is: . $\tfrac{5}{2} \;=\;\left(\tfrac{2}{5}\right)^{-1}$

The equation becomes: . $\left(\tfrac{2}{5}\right)^{2x} \;=\;\left(\tfrac{2}{5}\right)^{-1}$

Therefore: . $2x \:=\:-1 \quad\Rightarrow\quad x \:=\:-\tfrac{1}{2}$

$3)\;5(0.2)^{2-x} \:=\: \left(\tfrac{1}{25}\right)^{2x-1}$

The left side is: . $5(0.2)^{2-x} \;=\;5\left(\tfrac{1}{5}\right)^{2-x } \;=\;5\left(5^{-1}\right)^{2-x} \;=\;5\cdot 5^{x-2} \;=\;5^{x-1}$

The right side is: . $\left(\tfrac{1}{25}\right)^{2x-1} \;=\; \left(5^{-2}\right)^{2x-1} \;=\;5^{-4x+2}$

The equation becomes: . $5^{x-1} \;=\;5^{-4x+2}$

Therefore: . $x-1 \:=\:-4x+2 \quad\Rightarrow\quad 5x \:=\:3 \quad\Rightarrow\quad x \:=\:\tfrac{3}{5}$

$4)\;\left(\tfrac{1}{3}\right)^{2-x} - 6\left(9^{\frac{x-4}{2}}\right) +2\left(3^{x-6}\right) \:=\:29$

The first term is: . $\left(\tfrac{1}{3}\right)^{2-x} \;=\;\left(3^{-1}\right)^{2-x} \;=\;3^{x-2}$

The second term is: . $6\left(9^{\frac{x-4}{2}}\right) \;=\;6\left(3^2\right)^{\frac{x-4}{2}} \;=\;6\left(3^{x-4}\right) \;=\;2\cdot3\,(3^{x-4}) \;=\;2\cdot3^{x-3}$

The equation becomes: . $3^{x-2} - 2\cdot3^{x-3} + 2\cdot3^{x-6} - 29 \:=\:0$ . . . ugh!

I see that $x = 6$ is a solution.
Beyond that, you're on your own.

**MarkFL** · September 22nd 2012, 12:33 PM

If we continue where Soroban left off with the last one...

$3^{x-2}-\frac{2}{3}3^{x-2}+\frac{2}{81}3^{x-2}=29$

$\frac{29}{81}3^{x-2}=29$

$3^{x-2}=3^4$

$x-2=4$

$x=6$

**skaters** · September 23rd 2012, 03:31 AM

Thanks guys!

I did last one with substitution, it turned very ugly with big nummbers(still I managed

). Your method looks more suitable
$3^{x-2}-\frac{2}{3}3^{x-2}+\frac{2}{81}3^{x-2}=29$

$\frac{29}{81}3^{x-2}=29$
But I cant get what u do with $-\frac{2}{3}$

**MarkFL** · September 23rd 2012, 06:29 AM

$3^{x-2}-\frac{2}{3}3^{x-2}+\frac{2}{81}3^{x-2}=29$

Factor the left side:

$3^{x-2}\left(1-\frac{2}{3}+\frac{2}{81} \right)=29$

Get common denominator within parentheses:

$3^{x-2}\left(\frac{81-54+2}{81} \right)=29$

$\frac{29}{81}3^{x-2}=29$

$\frac{81}{29}\cdot\frac{29}{81}3^{x-2}=29\cdot\frac{81}{29}$

$3^{x-2}=81=3^4$

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