If ${log}_{y} 6 = x$ $Log_y6=x$ , ${log}_{y} 3.5 = z$ $Log_y3.5=z$ and $6 a = 3.5$ $6a=3.5$ then $a = ? ?$ $a=??$

Question

If ${log}_{y} 6 = x$ $Log_y6=x$ , ${log}_{y} 3.5 = z$ $Log_y3.5=z$ and $6 a = 3.5$ $6a=3.5$ then $a = ? ?$ $a=??$

$(a)$ $(a)$ $x z$ $xz$
$(b)$ $(b)$ $\frac{z}{x}$ $z/x$
$(c)$ $(c)$ $z + x$ $z+x$
$(d)$ $(d)$ $z - x$ $z-x$

I searched on internet about correct answer and it was $(b)$ $(b)$ $\frac{z}{x}$ $z/x$ , but I don't know why $(b)$ $(b)$ is correct. I think that all the answers listed are incorrect.

2329 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-15T07:42:09

I got $a = \frac{y^{z}}{y^{x}} = y^{z - x}$ $a=y^z/y^x=y^(z-x)$

$6 a = 3.5 \Rightarrow a = \frac{3.5}{6}$ $6a=3.5=>a=3.5/6$

We're told that

${log}_{y} 6 = x; {log}_{y} 3.5 = z$ $log_y6=x; log_y3.5=z$

We can rearrange these equations to put them into exponential form:

$y^{x} = 6; y^{z} = 3.5$ $y^x=6; y^z=3.5$

And substitute in the values:

$a = \frac{y^{z}}{y^{x}} = y^{z - x}$ $a=y^z/y^x=y^(z-x)$