Question

If $\log x+\log 5=\log {{x}^{2}}-\log 14$, the x is equal to
[a] ${{2}^{70}}$
[b] $70$
[c] 0 or 70
[d] ${{70}^{2}}$

Answer 1

Hint: Use the fact that $\log x$ is defined for x>0 to claim that the solution if exists is for x > 0. Use the fact that $\log {{x}^{2}}=2\log \left| x \right|$ and hence prove that the given equation is equivalent to solving $\log x=\log 5+\log 14$. Use the fact that $\log m+\log n=\log mn$ and hence find the value of x.

Complete step-by-step answer:

We know that $\log m+\log n=\log mn$
Using the above identity, we get
LHS $=\log x+\log 5=\log 5x$
We know that $\log m-\log n=\log \left( \dfrac{m}{n} \right)$
Using the above identity, we get
RHS $=\log {{x}^{2}}-\log 14=\log \left( \dfrac{{{x}^{2}}}{14} \right)$
Hence, we have
$\log 5x=\log \left( \dfrac{{{x}^{2}}}{14} \right)$
Since logx is a one-one function, we have
$5x=\dfrac{{{x}^{2}}}{14}$
Hence, we have $x=0$ or $x=5\times 14=70$
Since logx is defined for x>0, we have x=0 is rejected.
Hence, we have x= 70 is the solution of the given equation
Hence, option [b] is correct.

Note: Alternative Solution: Best method:
Since logx is defined for x>0, the solution exists for x>0 only.
Now, we know that $\log {{x}^{2}}=2\log \left| x \right|$
Since for x>0 |x|= x, we have $\log {{x}^{2}}=2\log x$
Now, we have $\log x+\log 5=2\log x-\log 14$
Adding log14-logx on both sides, we get
$\log 5+\log 14=\log x$
We know that $\log m+\log n=\log mn$
Hence, we have
$\log x=\log 70$
Since logx is a one-one function, we have
$x=70$, which is the same as obtained above.
Hence option [b] is correct.
[2] Verifiction:
log70+log5=log350
$\log {{70}^{2}}-\log 14=\log \dfrac{4900}{14}=\log 350$
Hence, we have $\log x+\log 5=\log {{x}^{2}}-\log 14$
Hence, our solution is verified to be correct.
[3] Many students forget to take into account the domain of logx and hence end up with 0 as one of the solutions which is incorrect.