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

[oded244]

?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

 

[galactus]

Change to base 10 and note the cancellations.

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

Change to base 10:

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

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

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

We get:

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

Now, solve for x.

 

[soroban]

Hello, oded244!

Another approach . . .

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

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

. . \(\displaystyle \log_5\left(2^{3x-6}\right)\cdot\log_2\left(5^{2x-2}\right) \;=\;2x+6\)

. . \(\displaystyle (3x-6)\!\cdot\!\log_5(2)\cdot(2x-2)\!\cdot\!\log_2(5) \;=\;2x+6\)

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

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

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