Question

Let S be the set of ordered triples $(x,y,z)$ of real number for which ${\log }_{10}(x+y)=z$ and ${\log }_{10}(x^2+y^2)=z+1$. If a,b are real number such that for all ordered triplets $(x,y,z)$ in S, we have $x^3+y^3=a.{10}^{3x}+b{.1}^{2x}$. Then the value of $9a+b)$ is

• A. $\frac{15}{2}$
• B. $\frac{29}{2}$
• C. $15$
• D. $\frac{39}{2}$

Answer 1

Answer: (B)

$\frac{29}{2}$

First, remember that $x^3+y^3$ factors to $(x+y)(x^2-xy+y^2)$.

By the givens, $x+y={10}^z$ and $x^2+y^2={10}^{z+1}$.

These can be used to find $xy$:

$(x+y{)}^2={10}^{2z}$

$x^2+2xy+y^2={10}^{2z}$

$2xy={10}^{2z}-{10}^{z+1}$

$xy=\frac{{10}^{2z}-{10}^{z+1}}{2}$

Therefore,

$x^3+y^3=a{.10}^{3z}+b.{10}^{2z}={10}^z({10}^{z+1}-\frac{{10}^{2z}-{10}^{z+1}}{2})$

$={10}^z({10}^{z+1}-\frac{{10}^{2z}-{10}^{z+1}}{2})$

$={10}^{2z+1}-\frac{{10}^{3z}-{10}^{2z+1}}{2}$

$=-\frac{1}{2}.{10}^{3z}+\frac{3}{2}.{10}^{2z+1}$

$=-\frac{1}{2}.{10}^{3z}+15{.10}^{2z}$

It follows that $a=-\frac{1}{2}$ and $b=15$,

thus $a+b=\frac{29}{2}.$