Question

How do you solve ${10^{7x}} = 76$?

Answer 1

Here we will use the logarithm on both the sides of the equation and will simplify the equation using the logarithmic table for the reference value and then find the value of the resultant required value, “x”.

Complete step-by-step solution:
Take the given expression,
${10^{7x}} = 76$
Take logarithm on both the sides of the equation.
$\log ({10^{7x}}) = \log 76$
Here we will use the Power rule: ${\log _a}{x^n} = n{\log _a}x$ in the above equation.
$7x.\log (10) = \log 76$
We know that $\log 10 = 1$so place it in the above equation.
$\Rightarrow 7x.(1) = \log 76$
Simplify the above equation,
$\Rightarrow 7x = \log 76$
Term multiplicative on one side, if moved to the opposite side, goes to the denominator.
$\Rightarrow x = \dfrac{{\log 76}}{7}$
Refer, natural logarithmic table for the reference value for log
$\log 76 = 1.8808$
Place the value in the above equation.
$\Rightarrow x = \dfrac{{\operatorname{l} .8808}}{7}$
Perform division,
$\Rightarrow x = 0.269$
The above equation can be re-written as –
$x \approx 3$

Additional Information: Also refer to the below properties and rules of the logarithm.
i) Product rule: ${\log _a}xy = {\log _a}x + {\log _a}y$
ii) Quotient rule: ${\log _a}\dfrac{x}{y} = {\log _a}x - {\log _a}y$
iii) Power rule: ${\log _a}{x^n} = n{\log _a}x$
iv) Base rule:${\log _a}a = 1$
v) Change of base rule: ${\log _a}M = \dfrac{{\log M}}{{\log N}}$

Note: Log table is used to find the value of the logarithmic function. Before the invention of the computers and other electronic devices, this log book was widely used and the easier way to get the required answer. In other words, the logarithm is the power to which the number must be raised in order to get some other. Always remember the standard properties of the logarithm.... Product rule, quotient rule and the power rule. The basic logarithm properties are most important and the solution solely depends on it, so remember and understand its application properly. Be good in multiples and know the concepts of square and square root and apply accordingly.