logarithm eqn: need help w/ log_x(64) + log_8x(32) = 4

[oded244]

need help...

 

[skeeter]

Re: logarithm ...

the typeset of the equation is not really clear ... is the equation

$\displaystyle \log(x^{64}) + \log(8x^{32}) = 4$

or

$\displaystyle \log_x(64) + \log_{8x}(32) = 4$

or something else?

 

[oded244]

Re: logarithm ...

im not sure, i just copied it from my book..
here's another example, i know you can change the bases to Log _2 but im wondering what i did wrong in my way.
(it's not related to the problem above)

 

[skeeter]

Re: logarithm ...

seems more likely that it's

$\displaystyle \log_x(64) + \log_{8x}(32) = 4$

change to base 2 ...

$\displaystyle \frac{\log_2(64)}{\log_2(x)} + \frac{\log_2(32)}{\log_2(8x)} = 4$

$\displaystyle \frac{6}{\log_2(x)} + \frac{5}{\log_2(8) + \log_2(x)} = 4$

$\displaystyle \frac{6}{\log_2(x)} + \frac{5}{3 + \log_2(x)} = 4$

let
$\displaystyle t = \log_2(x)$

$\displaystyle \frac{6}{t} + \frac{5}{3 + t} = 4$

$\displaystyle \frac{6(3 + t)}{t(3 + t)} + \frac{5t}{t(3 + t)} = 4$

$\displaystyle \frac{18+11t}{t(3+t)} = 4$

$\displaystyle 18+11t = 12t + 4t^2$

$\displaystyle 0 = 4t^2 + t - 18$

$\displaystyle 0 = (t-2)(4t+9)$

$\displaystyle t = 2$
$\displaystyle x = 4$

$\displaystyle t = -\frac{9}{4}$
$\displaystyle x = 2^{-\frac{9}{4}}$ ... throw this solution out, base x must be > 0.

 

[oded244]

Re: logarithm ...

Thanks skeeter.
About the other one, i know i can change to base 2 , but im trying to solve it in another way, can you see what im doing wrong?

 

[skeeter]

Re: logarithm ...

$\displaystyle \log_2(x) + \log_{2x}(x^3) = 4$

$\displaystyle \log_2(x) + \frac{\log_2(x^3)}{\log_2(2x)} = 4$

$\displaystyle \log_2(x) + \frac{3\log_2(x)}{\log_2(2) + \log_2(x)} = 4$

let
$\displaystyle t = \log_2(x)$

$\displaystyle t + \frac{3t}{1+t} = 4$

take it from here?

 

[oded244]

Re: logarithm ...

thanks skeeter for your help!
but im trying to solve it without changing the bases (see picture above) and trying to figure out what i did wrong.
Is it even possible to solve it without changing the bases to Log_2 ?

 

[soroban]

Re: logarithm ...

Hello, oded244!

$\displaystyle \log_x(64) + \log_{8x}(32) \;=\;4$

$\displaystyle \text{We have: }\;\log_x(2^6) + \log_{8x}(2^5) \;=\;4 \quad\Rightarrow\quad 6\log_x(2) + 5\log_{8x}(2) \;=\;4$

. . $\displaystyle \frac{6}{\log_2(x)} + \frac{5}{\log_2(8x)} \;=\;4 \quad\Rightarrow\quad 6\log_2(8x) + 5\log_2(x) \;=\;4\log_2(x)\log_2(8x)$

. . $\displaystyle 6\bigg[\log_2(8) + \log_2(x)\bigg] + 5\log_2(x) \;=\;4\log_2(x)\bigg[\log_2(8) + \log_2(x)\bigg]$

. . $\displaystyle 6\cdot 3 + 6\log_2(x) + 5\log_2(x) \;=\;4\log_2(x)\bigg[3 + \log_2(x)\bigg]$

. . $\displaystyle 18 + 11\log_2(x) \;=\;12\log_2(x) + 4\left[\log_2(x)\right]^2$


$\displaystyle \text{We have the quadratic: }\;4\left[\log_2(x)\right]^2 + \log_2(x) - 18 \;=\;0$

. . $\displaystyle \text{which factors: }\;\left[4\log_2(x) + 9\right]\left[\log_2(x) - 2\right] \;=\;0$


Therefore . . .

$\displaystyle 4\log_2(x) + 9 \:=\:0 \quad\Rightarrow\quad \log_2(x) \:=\:-\frac{9}{4} \quad\Rightarrow\quad\boxed{ x \;=\;2^{-\frac{9}{4}}}$

$\displaystyle \log_2(x) - 2 \:=\:0 \quad\Rightarrow\quad \log_2(x) \:=\:2 \quad\Rightarrow\quad x \:=\:2^2 \quad\Rightarrow\quad\boxed{x\:=\:4}$

 

[oded244]

Re: logarithm ...

Great forum!
Thanks all.

 

[soroban]

Re: logarithm ...

Hello again, oded244!

$\displaystyle \log_2(x) + \log_{2x}(x^3) \;=\;4$

$\displaystyle \text{We have: }\;\log_2(x) + 3\log_{2x}(x) \;=\;4$

. . $\displaystyle \frac{1}{\log_x(2)} + \frac{3}{\log_x(2x)} \;=\;4$

. . $\displaystyle \log_x(2x) + 3\log_x(2) \;=\;4\log_x(2)\log_x(2x)$

. . $\displaystyle \log_x(2) + \log_x(x) + 3\log_x(2) \;=\;4\log_x(2)\bigg[\log_x(2) + \log_x(x)\bigg]$

. . $\displaystyle 4\log_x(2) + 1 \;=\;4\log_x(2)\bigg[\log_x(2) + 1\bigg]$

. . $\displaystyle 4\log_x(2) + 1 \;=\;4\bigg[\log_x(2)\bigg]^2 + 4\log_x(2)$


$\displaystyle \text{We have: }\;4\bigg[\log_x(2)\bigg]^2 \;=\;1 \quad\Rightarrow\quad\bigg[\log_x(2)\bigg]^2 \;=\;\frac{1}{4} \quad\Rightarrow\quad \log_x(2) \;=\;\pm\frac{1}{2}$

. . $\displaystyle \log_x(2) \:=\:\frac{1}{2} \quad\Rightarrow\quad x^{\frac{1}{2}} \:=\:2 \quad\Rightarrow\quad\boxed{ x \:=\:4}$

. . $\displaystyle \log_x(2) \:=\:-\frac{1}{2} \quad\Rightarrow\quad x^{-\frac{1}{2}} \:=\:2 \quad\Rightarrow\quad \boxed{x \:=\:\frac{1}{4}}$