Solve Logarithm Problem: Exponential Form & Single Expression

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To solve the equation Log2(2-5x)=7, it can be rewritten in exponential form as 2^7 = 2 - 5x, leading to the solution for x. For the second part, the expression 3log x + (1/2)log z can be combined into a single logarithmic expression using the rules of logarithms, resulting in log(x^3 * z^(1/2)). Understanding that logarithms represent exponents helps clarify these transformations. The discussion emphasizes the importance of using logarithmic properties to simplify and solve equations effectively.
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Homework Statement


1) Solve the equation using the exponential form of the equation.

Log2(2-5x)=7

2) Rewrite as a single logarithmic expression:

3log x +(1/2)log z

Homework Equations


The Attempt at a Solution



1) I have no idea how to solve this.

2) I know that two logarithms multiply to add, but these have different bases so I do not know what to do.

Thanks.
 
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math4life said:
Log2(2-5x)=7

Here is a hint.

a^b =c \Rightarrow b= log_a c

math4life said:
2) Rewrite as a single logarithmic expression:

3log x +(1/2)log z

they are in the same base. Logx usually means log10x.

Use the rule

rlog_a x = log_a x^2
 
I'll just fix up rock.freak667's little typo there.

What was meant to be said is: r.log_a(x)=log_a(x^r)

Coupled with the rule that log_a(b)+log_a(c)=log_a(bc)

You should have no problem solving the second one :smile:
 
I think it might be helpful to add that a logarithm can be thought of as the exponent on the base (the number raised to a power) that results in a particular number. So for example, log10100 means the exponent on 10 that results in 100. In other words, this logarithm is the answer to the question 10? = 100. Pretty clearly, the placeholder represented by ? is 2.

Every equation of the form logbx = y can be rewritten as an equivalent exponential equation by = x, and vice versa, with the only restrictions being that b > 0, and b\neq 1, and x > 0.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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