Inverting a Log Function: Using multiple rules

Examples with solutions for Inverting a Log Function: Using multiple rules

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

2log78log74+1log43×log29= \frac{2\log_78}{\log_74}+\frac{1}{\log_43}\times\log_29=

Video Solution

Answer

7 7

Exercise #3

log311log34+1ln32log3= \frac{\log_311}{\log_34}+\frac{1}{\ln3}\cdot2\log3=

Video Solution

Answer

log411+loge2 \log_411+\log e^2

Exercise #4

log76log71.53log721log82= \frac{\log_76-\log_71.5}{3\log_72}\cdot\frac{1}{\log_{\sqrt{8}}2}=

Video Solution

Answer

1 1

Exercise #5

3(ln4ln5log57+1log65)= -3(\frac{\ln4}{\ln5}-\log_57+\frac{1}{\log_65})=

Video Solution

Answer

3log5724 3\log_5\frac{7}{24}

Exercise #6

1ln41log810= \frac{1}{\ln4}\cdot\frac{1}{\log_810}=

Video Solution

Answer

32loge \frac{3}{2}\log e

Exercise #7

log8x3log8x1.5+1log49x×log7x5= \frac{\log_8x^3}{\log_8x^{1.5}}+\frac{1}{\log_{49}x}\times\log_7x^5=

Video Solution

Answer

12 12

Exercise #8

1logx3×x2log1x27+4x+6=0 \frac{1}{\log_x3}\times x^2\log_{\frac{1}{x}}27+4x+6=0

x=? x=\text{?}

Video Solution

Answer

23+223 \frac{2}{3}+\frac{\sqrt{22}}{3}

Exercise #9

1log2x6×log236=log5(x+5)log52 \frac{1}{\log_{2x}6}\times\log_236=\frac{\log_5(x+5)}{\log_52}

x=? x=\text{?}

Video Solution

Answer

1.25 1.25

Exercise #10

Find X

1logx42×xlogx16+4x2=7x+2 \frac{1}{\log_{x^4}2}\times x\log_x16+4x^2=7x+2

Video Solution

Answer

9+1138 \frac{-9+\sqrt{113}}{8}