Find the value of $y+z$, if $3^{\log_a x}+3x^{\log_a 3}=2$ and $x=y^{-\log_z a}$.

Question

Find the value of $y+z$, if $3^{\log_a x}+3x^{\log_a 3}=2$ and $x=y^{-\log_z a}$

My tries:

$$3^{\log_a x}=x^{\log_a 3}$$ $$\text{Taking } 3^{\log_a x}=x^{\log_a 3}=p,$$ $$p+3p=2$$ $$or,p=1/2$$

Answer 1

$$x^{\log_{a}3} = 0.5 \implies 3^{\log_a x} = 0.5 \implies \log_a x = \log_3 0.5$$

$$x = a^{-\log_z y} \implies \log_a x = -\log_z y \implies \log_zy = -\log_3 0.5 = \log_32$$

$$\log_zy = \log_32 \implies y+z = 5$$

Answer 2

Well, we can use:

$$\text{a}^{\log_\text{b}\left(\text{c}\right)}=\text{c}^{\log_\text{b}\left(\text{a}\right)}\tag1$$

So, we get:

$$3^{\log_\text{a}\left(x\right)}+3x^{\log_\text{a}\left(3\right)}=3^{\log_\text{a}\left(x\right)}+3^1\cdot3^{\log_\text{a}\left(x\right)}=3^{\log_\text{a}\left(x\right)}+3^{1+\log_\text{a}\left(x\right)}=2\tag2$$

Substitute $\text{u}:=\log_\text{a}\left(x\right)$ and after that substitute $\text{q}:=3^\text{u}$:

$$3^\text{u}+3^{1+\text{u}}=2\space\Longleftrightarrow\space4\cdot\text{q}=2\space\Longleftrightarrow\space\text{q}=3^\text{u}=\frac{1}{2}\space\Longleftrightarrow$$ $$\text{u}=\log_\text{a}\left(x\right)=-\log_3\left(2\right)=\log_3\left(\frac{1}{2}\right)\tag3$$

And we also know that:

$$x=\text{y}^{-\log_\text{z}\left(\text{a}\right)}=\text{y}^{\log_\text{z}\left(\frac{1}{\text{a}}\right)}=\left(\frac{1}{\text{a}}\right)^{\log_\text{z}\left(\text{y}\right)}\tag4$$

So, we get:

$$\log_\text{a}\left(\left(\frac{1}{\text{a}}\right)^{\log_\text{z}\left(\text{y}\right)}\right)=\log_\text{z}\left(\text{y}\right)\cdot\log_\text{a}\left(\frac{1}{\text{a}}\right)=-\log_\text{z}\left(\text{y}\right)=-\log_3\left(2\right)\tag5$$