# Thread: 3 examples

1. ## 3 examples

I cannot solve these 3 examples. If you know it, please help me.

2. To simplify calculations,
for (1), begin by factorising 35 to write it as a multiple of the single-digit terms, 5 and 7, may be simpler ways though.

For (2), 9 is a power of 3, 4 is a power of 2 and 6 is a multiple of 3 and 2.

For (3), 4 and 8 are powers of 2, 12 is a multiple of 2, but this is a quadratic in disguise.

3. I understand what you said but what to do after it?

4. Well, for (1)

simplest is to write $5^x+(35)7^x=7^x+(35)5^{2x}$

If x=0, then

$5^0+(35)7^0=7^0+(35)5^0$

$1+35=1+35$

However, I doubt that this is the way you are expected to solve,
since this answer is by inspection,
it is not a derived solution.

5. $5^x+(35)7^x=7^x+(35)5^{2x}$

$\frac{5^x}{35}-5^{2x}=\frac{7^x}{35}-7^x$

$5^{2x}(\frac{5^{-x}}{35}-1)=7^x(\frac{1}{35}-1)$

These are equal if x=0

6. The third is really simple.
Don't forget to find a domain of the function.
$2^{2(x- \sqrt{x^2 - 5})} - 12 \cdot 2^{x- \sqrt{x^2 - 5}} + 8 = 0$

$2^{x- \sqrt{x^2 - 5}}=a$ and a>0
$a^2 - 12a + 8 = 0$

7. For Q2

$(6)4^{x+1}+\frac{9^{x+1}}{2}=\frac{9^{x+2}}{3}+(3) 4^x$

$(6)2^{2(x+1)}+\frac{3^{2(x+1)}}{2}=3^{2(x+2)-1}+(3)2^{2x}$

$(6)(4)2^{2x}+\frac{3^{2x+3}}{6}=3^{2x+3}+(3)2^{2x}$

continue

8. 2. Mulitply two sides of the equation throughout 6.
After simple transformation:
$36 \cdot 4^{x+1}=2 \cdot 9^{x+2} - 3 \cdot 9^{x+1} +18 \cdot 4^x$
$9 \cdot 4^{x+2} = 2 \cdot 9^{x+2} - 3 \cdot 9^{x+1} + 3 \cdot 4^x$
$4^x(4^2 \cdot 9 - 18) = 9^x(2 \cdot 9^2 - 3 \cdot 9)$
$(\frac{2}{3})^x = \frac{135}{126}$

If I didn't make a mistake to give a result you need to use a logarithm.

9. Hello, x-mather!

These are truly ugly problems . . .

$2)\;\;6\cdot4^{x+1} \;=\;\frac{9^{x+2}}{3} - \frac{9^{x+1}}{2} + 3\cdot4^x$

We have: . $2\cdot3\cdot(2^2)^{x+1} \;=\;\frac{(3^2)^{x+2}}{3} - \frac{(3^2)^{x+1}}{2} + 3\cdot (2^2)^x$

. . . . . . . . . $2\cdot3\cdot2^{2x+2} \;=\;\frac{3^{2x+4}}{3} - \frac{3^{2x+2}}{2} + 3\cdot2^{2x}$

. . . . . . . . . . $3\cdot2^{2x+3} \;=\;3^{2x+3} - \frac{3^{2x+2}}{2} + 3\cdot2^{2x}$

Multiply by 2: . $3\cdot2^{2x+4} \;=\;2\cdot3^{2x+3} - 3^{2x+2} + 3\cdot2^{2x+1}$

. . . . $3\cdot2^{2x+4} - 3\cdot2^{2x+1} \;=\;2\cdot3^{2x+3} - 3^{2x+2}$

Factor: . $3\cdot2^{2x+1}(2^3-1) \;=\;3^{2x+2}(2\cdot3-1)$

. . . . . . . . . $3\cdot2^{2x+1}\cdot7 \;=\;3^{2x+1}\cdot5$

Divide by 3: . . $7\cdot2^{2x+1} \;=\;5\cdot3^{2x}$

. . . . . . . . . . . . $\frac{2^{2x+1}}{3^{2x}} \;=\;\frac{5}{7}$

Divide by 2: . . . . $\frac{2^{2x}}{3^{2x}} \;=\;\frac{5}{14}$

. . . . . . . . . . . $\left(\frac{2}{3}\right)^{2x} \;=\;\frac{5}{14}$

Take logs: . . $\ln\left(\tfrac{2}{3}\right)^{2x} \;=\;\ln\left(\tfrac{5}{14}\right)$

. . . . . . . . . $2x\ln\left(\tfrac{2}{3}\right) \;=\;\ln\left(\tfrac{5}{14}\right)$

. . Therefore: . . . . $x \;=\;\frac{\ln\left(\frac{5}{14}\right)}{2\ln\left (\frac{2}{3}\right)}$

10. Late, while I was writing I didn't see a new post appeared.