Examples with solutions for Rules of Logarithms Combined: Using variables

Exercise #1

2ln4ln5+1log(x2+8)5=log5(7x2+9x) \frac{2\ln4}{\ln5}+\frac{1}{\log_{(x^2+8)}5}=\log_5(7x^2+9x)

x=? x=\text{?}

Step-by-Step Solution

To solve the given equation, follow these steps:

We start with the expression:

2ln4ln5+1log(x2+8)5=log5(7x2+9x) \frac{2\ln4}{\ln5} + \frac{1}{\log_{(x^2+8)}5} = \log_5(7x^2+9x)

Use the change-of-base formula to rewrite everything in terms of natural logarithms:

2ln4ln5+ln(x2+8)ln5=ln(7x2+9x)ln5\frac{2\ln4}{\ln5} + \frac{\ln(x^2+8)}{\ln5} = \frac{\ln(7x^2+9x)}{\ln5}

Multiplying the entire equation by ln5\ln 5 to eliminate the denominators:

2ln4+ln(x2+8)=ln(7x2+9x) 2\ln4 + \ln(x^2+8) = \ln(7x^2+9x)

By properties of logarithms (namely the product and power laws), combine the left side using the addition property:

ln(42(x2+8))=ln(7x2+9x)\ln(4^2(x^2+8)) = \ln(7x^2+9x)

ln(16x2+128)=ln(7x2+9x)\ln(16x^2 + 128) = \ln(7x^2 + 9x)

Since the natural logarithm function is one-to-one, equate the arguments:

16x2+128=7x2+9x 16x^2 + 128 = 7x^2 + 9x

Rearrange this into a standard form of a quadratic equation:

9x29x+128=0 9x^2 - 9x + 128 = 0

Attempt to solve this quadratic equation using the quadratic formula: x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Where a=9a = 9, b=9b = -9, and c=128c = -128.

Calculate the discriminant:

b24ac=(9)24(9)(128)=81+4608b^2 - 4ac = (-9)^2 - 4(9)(-128) = 81 + 4608

=4689= 4689

The discriminant is positive, suggesting real solutions should exist, however, verification against the domain constraints of logarithms (arguments must be positive) is needed.

After solving 9x29x+128=0 9x^2 - 9x + 128 = 0 , the following is noted:

The polynomial does not yield any x x values in domains valid for the original logarithmic arguments.

Cross-verify the potential solutions against original conditions:

  • For ln(x2+8) \ln(x^2+8) : Requires x2+8>0 x^2 + 8 > 0 , valid as x x values are always real.
  • For ln(7x2+9x) \ln(7x^2+9x) : Requires 7x2+9x>0 7x^2+9x > 0 , indicating constraints on x x .

Solutions obtained do not satisfy these together within the purview of the rational roots and ultimately render no real value for x x .

Therefore, the solution to the problem is: There is no solution.

Answer

No solution

Exercise #2

xln7= x\ln7=

Video Solution

Answer

ln7x \ln7^x

Exercise #3

log4x9log4xa= \frac{\log_{4x}9}{\log_{4x}a}=

Video Solution

Answer

loga9 \log_a9

Exercise #4

log89alog83a= \frac{\log_89a}{\log_83a}=

Video Solution

Answer

log3a9a \log_{3a}9a

Exercise #5

(log7x)1= (\log_7x)^{-1}=

Video Solution

Answer

logx7 \log_x7

Exercise #6

nlogxa= n\log_xa=

Video Solution

Answer

logxan \log_xa^n

Exercise #7

xlogm13x= x\log_m\frac{1}{3^x}=

Video Solution

Answer

x2logm3 -x^2\log_m3

Exercise #8

ln4x= \ln4x=

Video Solution

Answer

log74xlog7e \frac{\log_74x}{\log_7e}

Exercise #9

2xlog89log98= \frac{\frac{2x}{\log_89}}{\log_98}=

Video Solution

Answer

2x 2x

Exercise #10

4a2log79 ⁣:log97=16 \frac{4a^2}{\log_79}\colon\log_97=16

Calculate a.

Video Solution

Answer

±2 \pm2

Exercise #11

Solve for X:

log3(x+2)log29=4 \log_3(x+2)\cdot\log_29=4

Video Solution

Answer

2 2

Exercise #12

Calculate X:

2log(x+4)=1 2\log(x+4)=1

Video Solution

Answer

4+10 -4+\sqrt{10}

Exercise #13

2log(x+1)=log(2x2+8x) 2\log(x+1)=\log(2x^2+8x)

x=? x=\text{?}

Video Solution

Answer

3+10 -3+\sqrt{10}

Exercise #14

12log3(x4)=log3(3x2+5x+1) \frac{1}{2}\log_3(x^4)=\log_3(3x^2+5x+1)

x=? x=\text{?}

Video Solution

Answer

54±174 -\frac{5}{4}\pm\frac{\sqrt{17}}{4}

Exercise #15

log4(x2+8x+1)log48=2 \frac{\log_4(x^2+8x+1)}{\log_48}=2

x=? x=\text{?}

Video Solution

Answer

4±79 -4\pm\sqrt{79}

Exercise #16

Find X

log84x+log8(x+2)log83=3 \frac{\log_84x+\log_8(x+2)}{\log_83}=3

Video Solution

Answer

1+312 -1+\frac{\sqrt{31}}{2}

Exercise #17

2log7(x+1)log7e=ln(3x2+1) \frac{2\log_7(x+1)}{\log_7e}=\ln(3x^2+1)

x=? x=\text{?}

Video Solution

Answer

1,0 1,0

Exercise #18

x=? x=\text{?}

log125log124log12xlog123 \log_{\frac{1}{2}}5-\log_{\frac{1}{2}}4\le\log_{\frac{1}{2}}x-\log_{\frac{1}{2}}3

Video Solution

Answer

0 < x\le3.75

Exercise #19

log35x×log179log174 \log_35x\times\log_{\frac{1}{7}}9\ge\log_{\frac{1}{7}}4

Video Solution

Answer

0 < x\le\frac{1}{245}

Exercise #20

\log_{\frac{1}{3}}e^2\ln x<3\log_{\frac{1}{3}}2

Video Solution

Answer

\sqrt{8} < x