Question

Solve the following equation for x: $\frac{5}{a}$ loga x.log10 a. loga 5 - 3log10(x/10) = 9log100x+log4 2

Answer 1

$x = 10^{5/13}$

Answer 2

The given equation is $\frac{5}{a}$ loga x.log10 a. loga 5 - 3log10(x/10) = 9log100x+log4 2.

Using the change of base formula for logarithms ($log_b c = \frac{log_d c}{log_d b}$), we simplify the terms.

The first term: $\frac{5}{a} \cdot log_a x \cdot log_{10} a \cdot log_a 5$. We know that $log_a x \cdot log_{10} a = \frac{log_{10} x}{log_{10} a} \cdot log_{10} a = log_{10} x$. So the first term becomes $\frac{5}{a} \cdot log_{10} x \cdot log_a 5$.

For simplification and based on the structure of similar problems, we assume $a=5$. If $a=5$, then $log_a 5 = log_5 5 = 1$. The first term simplifies to $\frac{5}{5} \cdot log_{10} x \cdot 1 = log_{10} x$.

Now, let's simplify the other terms: $-3log_{10}(x/10) = -3(log_{10} x - log_{10} 10) = -3(log_{10} x - 1) = -3log_{10} x + 3$. $9log_{100} x = 9 \cdot \frac{log_{10} x}{log_{10} 100} = 9 \cdot \frac{log_{10} x}{2} = \frac{9}{2} log_{10} x$. $log_4 2 = \frac{1}{2}$ (since $4^{1/2} = 2$).

Substitute these simplified terms back into the equation (with $a=5$): $log_{10} x + (-3log_{10} x + 3) = \frac{9}{2} log_{10} x + \frac{1}{2}$.

Let $Y = log_{10} x$. The equation becomes: $Y - 3Y + 3 = \frac{9}{2} Y + \frac{1}{2}$. $-2Y + 3 = \frac{9}{2} Y + \frac{1}{2}$.

Rearrange the terms to solve for $Y$: $3 - \frac{1}{2} = \frac{9}{2} Y + 2Y$. $\frac{6-1}{2} = \frac{9Y+4Y}{2}$. $\frac{5}{2} = \frac{13Y}{2}$. $5 = 13Y$. $Y = \frac{5}{13}$.

Since $Y = log_{10} x$, we have $log_{10} x = \frac{5}{13}$. Therefore, $x = 10^{5/13}$.