Question

Solve the following equations, having given log2, log3 and log7.
    ${3^{1 - x - y}} = {4^{ - y}}$
     ${2^{2x - 1}} = {3^{3y - 1}}$

Answer 1

Hint- Make use of the formula of product rule, division rule and power rule of logarithms and try to solve this problem

Complete step-by-step answer:

Let us consider the first equation ${3^{1 - x - y}} = {4^{ - y}}$

To solve this, let us apply log on both the sides that is on LHS and RHS

So, we get $\log ({3^{1 - x - y}}) = \log ({4^{ - y}})$

Now , let us make use of the power rule of log and try to solve this which says ${\log _a}{m^n} = n{\log _a}m$

So, on applying this rule in the equation, we get

(1-x-y)log3=-ylog4

(1-x-y)log3=-ylog${2^2}$

(1-x-y)log3=-2ylog2

Let’s make use of the formula of $\dfrac{{\log m}}{{\log n}} = \log m - \log n$ and try to find out the value of y

On solving further and shifting the terms, we get

$y = \dfrac{{\log 3(1 - x)}}{{\log 3 - 2\log 2}}$ --------------(i)

Now, let us take the second equation ${2^{2x - 1}} = {3^{3y - 1}}$

Let us apply log on both the sides

So, we get $\log ({2^{2x - 1}}) = \log ({3^{3y - 1}})$

Now , let us make use of the formula ${\log _a}{m^n} = n{\log _a}m$

So, we get the equation as (2x-1)log2=(3y-1)log3

2xlog2-log2=3ylog3-log3---------(ii)

From eq(i) we already had the value of $y = \dfrac{{\log 3(1 - x)}}{{\log 3 - 2\log 2}}$

So, let us substitute this value in eq(ii)

So, we get $2x\log 2 - \log 2 = 3\log 3\dfrac{{\log 3(1 - x)}}{{\log 3 - \log 2}} - \log 3$

 Making use of the product rule of logarithm which says ${\log _a}mn = {\log _a}m + {\log _a}n$

 Division rule of logarithm which says ${\log _a}\dfrac{m}{n} = {\log _a}m - {\log _a}n$

And power rule of logarithm which says ${\log _a}{m^n} = n{\log _a}m$

We get x=$\dfrac{{3{{(\log 3)}^2} + 3\log 3\log 2 - 2{{(\log 2)}^2}}}{{3{{(\log 3)}^2} + 2\log

2\log 3 - 4{{(\log 2)}^2}}}$

$y = \dfrac{{\log 2\log 3}}{{3{{(\log 3)}^2} + 2\log 2\log 3 - 4{{(\log 2)}^2}}}$

Note: The logarithm of the exponent of x raised to the power of y, is y times the logarithm of x. The power rule can be used for fast exponent calculation using multiplication operation. Make use of the appropriate formula(product, division, power rule) of logarithms wherever needed and try to solve the problem.