need help - ?log?_5 8^(x-2)*?log?_2 ?25?^(x-1)=2x+6

oded244

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?log?_5 8^(x-2)*?log?_2 ?25?^(x-1)=2x+6

I've tried to change to base to Log2 but im getting stuck half way trough,

Thanks
 
Change to base 10 and note the cancellations.

log5(8x2)log2(25x1)=2x+6\displaystyle log_{5}(8^{x-2})\cdot log_{2}(25^{x-1})=2x+6

Change to base 10:

log(8x2)log(5)log(25x1)log(2)=2x+6\displaystyle \frac{log(8^{x-2})}{log(5)}\cdot \frac{log(25^{x-1})}{log(2)}=2x+6

=3(x2)/////log(2)/////log(5)2(x1)/////log(5)/////log(2)=2x+6\displaystyle =\frac{3(x-2)\rlap{/////}log(2)}{\rlap{/////}log(5)}\cdot \frac{2(x-1)\rlap{/////}log(5)}{\rlap{/////}log(2)}=2x+6

See, the log(5) and log(2) cancel.

We get:

3(x2)2(x1)=2x+6\displaystyle 3(x-2)\cdot 2(x-1)=2x+6

Now, solve for x.
 
Hello, oded244!

Another approach . . .


log5 ⁣(8x2)log2 ⁣(25x1)  =  2x+6\displaystyle \log_5\!\left(8^{x-2}\right)\cdot\log_2\!\left(25^{x-1}\right) \;=\;2x+6

We have:   log5(23)x2log2(52)x1  =  2x+6\displaystyle \text{We have: }\;\log_5\left(2^3\right)^{x-2}\cdot\log_2\left(5^2\right)^{x-1} \;=\;2x+6

. . log5(23x6)log2(52x2)  =  2x+6\displaystyle \log_5\left(2^{3x-6}\right)\cdot\log_2\left(5^{2x-2}\right) \;=\;2x+6

. . (3x6) ⁣ ⁣log5(2)(2x2) ⁣ ⁣log2(5)  =  2x+6\displaystyle (3x-6)\!\cdot\!\log_5(2)\cdot(2x-2)\!\cdot\!\log_2(5) \;=\;2x+6

. . 2(3x6)(x1)log5(2) ⁣ ⁣log2(5)This is 1  =  2(x+3)\displaystyle 2(3x-6)(x-1)\cdot\underbrace{\log_5(2)\!\cdot\!\log_2(5)}_{\text{This is 1}} \;=\;2(x + 3)

. . (3x6)(x1)  =  x+3\displaystyle (3x-6)(x-1) \;=\;x+3

Therefore:   3x210x+3=0(x3)(3x1)=0x=3,13\displaystyle \text{Therefore: }\;3x^2-10x+3 \:=\:0 \quad\Rightarrow\quad (x - 3)(3x-1) \:=\:0 \quad\Rightarrow\quad\boxed{ x \:=\:3,\:\frac{1}{3}}

 
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