log7x4−log72x2=3
?=x
\( \log_7x^4-\log_72x^2=3 \)
?=x
\( \log_89-\log_83+\log_4x^2=\log_81.5+\log_82+\log_4(-x^2-11x-9) \)
?=x
\( \frac{2\ln4}{\ln5}+\frac{1}{\log_{(x^2+8)}5}=\log_5(7x^2+9x) \)
\( x=\text{?} \)
\( \log_2x+\log_2\frac{x}{2}=5 \)
?=x
\( \log_4x^2\cdot\log_716=2\log_78 \)
?=x
?=x
We multiply by:
Extract the root
?=x
To solve the equation: , we proceed as follows:
Step 1: Simplify Both Sides
On the left-hand side (LHS), apply logarithmic subtraction:
.
Note remains and convert using the base switch to :
.
Thus, the LHS combines into:
(because ).
On the right-hand side (RHS):
Combine:
.
Also apply for term:
.
Step 2: Equalize Both Sides
Equate LHS and RHS logarithmic expressions:
.
The cancels out on both sides, leaving:
.
Step 3: Solve for
Since the denominators are equal, set the numerators equal:
.
Translate this into an exponential equation:
or
.
Let , solve the resulting quadratic equation:
.
Then, finding valid by allowing roots of polynomial calculations should yield laws consistency:
or rather substituting potential values. After appropriate checks:
The valid that satisfies the problem is thus .
To solve the given equation, follow these steps:
We start with the expression:
Use the change-of-base formula to rewrite everything in terms of natural logarithms:
Multiplying the entire equation by to eliminate the denominators:
By properties of logarithms (namely the product and power laws), combine the left side using the addition property:
Since the natural logarithm function is one-to-one, equate the arguments:
Rearrange this into a standard form of a quadratic equation:
Attempt to solve this quadratic equation using the quadratic formula:
Where , , and .
Calculate the discriminant:
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 , the following is noted:
The polynomial does not yield any values in domains valid for the original logarithmic arguments.
Cross-verify the potential solutions against original conditions:
Solutions obtained do not satisfy these together within the purview of the rational roots and ultimately render no real value for .
Therefore, the solution to the problem is: There is no solution.
No solution
?=x
?=x
Solve for X:
\( \log_3(x+2)\cdot\log_29=4 \)
\( \log_4x+\log_4(x+2)=2 \)
?=a
\( \ln(a+5)+\ln(a+7)=0 \)
\( \log3x+\log(x-1)=3 \)
\( ?=x \)
\( \ln(4x+3)-\ln(x^2-8)=2 \)
?=x
Solve for X:
?=a
?=x
\( \log7\times\ln x=\ln7\cdot\log(x^2+8x-8) \)
?=x
\( \log_27\cdot\log_48\cdot\log_3x^2=\log_24\cdot\log_47\cdot\log_38 \)
?=x
\( \log_2(x^2+3x+3)\cdot\log_3\frac{1}{4}=-2\log_3(\frac{4x+2}{-2}) \)
?=x
Calculate X:
\( 2\log(x+4)=1 \)
\( 2\log(x+1)=\log(2x^2+8x) \)
\( x=\text{?} \)
?=x
?=x
?=x
Calculate X:
\( \frac{1}{2}\log_3(x^4)=\log_3(3x^2+5x+1) \)
\( x=\text{?} \)
\( \frac{\log_4(x^2+8x+1)}{\log_48}=2 \)
\( x=\text{?} \)
Find X
\( \frac{\log_84x+\log_8(x+2)}{\log_83}=3 \)
\( \frac{2\log_7(x+1)}{\log_7e}=\ln(3x^2+1) \)
\( x=\text{?} \)
\( \log_4(3x^2+8x-10)-\log_4(-x^2-x+12.5)=0 \)
?=x
Find X
?=x