Solving Log expressions

**bharriga** · April 21st 2008, 11:15 AM

i am lost on what to do with these two questions:
they are more intense than other ones we have even looked at in class

solve for x:

8^6x-6 = (1/16)^8x-2
x ='s ___ ??

and another similar one:

2^2x+9 =3^x-44
x='s ___ in terms of logarithms or to four decimal places.

any guidance would be great!

**Moo** · April 21st 2008, 11:20 AM

Hello,

Is it

$8^{6x}-6 = (1/16)^{8x}-2$

Or

$8^{6x-6} = (1/16)^{8x-2}$

?

$2^{2x}+9 =3^x-44$

Or

$2^{2x+9} =3^{x-44}$

?

**Soroban** · April 21st 2008, 11:51 AM

Hello, bharriga!

You should use parentheses when writing math.
I'll have to guess what you meant.

$8^{6x-6} \:= \:\left(\frac{1}{16}\right)^{8x-2}$

We have: . $\left(2^3\right)^{6x-6} \:=\:\left(2^{-4}\right)^{8x-2} \quad\Rightarrow\quad 2^{18x-18} \:=\:2^{-3x+8}$

Hence: . $18x-18 \:=\:-32x+8\quad\Rightarrow\quad 50x \:=\:26\quad\Rightarrow\quad\boxed{ x \:=\:\frac{13}{25}}$

$2^{2x+9} \:=\:3^{x-44}$

Take logs: . $\ln\left(2^{2x+9}\right) \:=\:\ln\left(3^{x-44}\right) \quad\Rightarrow\quad (2x+9)\!\cdot\!\ln(2) \:=\:(x-44)\!\cdot\!\ln(3)$

. . $2x\cdot\ln(2) + 9\cdot\ln(2) \:=\:x\cdot\ln(3) - 44\cdot\ln(3) \quad\Rightarrow\quad x\cdot\ln(3) - 2x\cdot\ln(2) \:=\:9\cdot\ln(2) - 44\cdot\ln(3)$

Factor: . $x\,[\ln(3) - 2\!\cdot\!\ln(2)] \:=\:9\!\cdot\!\ln(2) - 44\!\cdot\!\ln(3)$

Therefore . $\boxed{x \;=\;\frac{9\!\cdot\!\ln(2) - 44\!\cdot\!\ln(3)}{\ln(3) - 2\!\cdot\!\ln(2)} \;\approx\;-189.7138}$

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