# Thread: Exponents and Logarithms-- 3 questions.

1. ## Exponents and Logarithms-- 3 questions.

Hi there i dont know if i should make separate threads maybe some one could let me know if so but here are three question i wanted to check if i was right or wrong without further adieu

1) Graph the following function: $\displaystyle f(x)=-\log(x+2)-1$ when i usually see a log function graph question i change to exponential form and when i do i get $\displaystyle 10^1=-(x+2)$ should it be $\displaystyle y+1=-(x+2)$ ?

2)Solve the following question leave in exact form: $\displaystyle (6^{3x+1})=(2^{2x-3})$ which i turn into $\displaystyle 3x+1\log(6)=2x-3\log(2)$ then i factor it out and bring x's together and solve from there i get $\displaystyle x= 1/24$

2. ## Re: Exponents and Logarithms-- 3 questions.

Let us see what have you tried so far.

3. ## Re: Exponents and Logarithms-- 3 questions.

Originally Posted by Gurp925
2)Solve the following question leave in exact form: $\displaystyle (6^{3x+1})=(2^{2x-3})$ which i turn into $\displaystyle 3x+1\log(6)=2x-3\log(2)$ then i factor it out and bring x's together and solve from there i get $\displaystyle x= 1/24$
No, this should be \displaystyle \displaystyle \begin{align*} \left( 3x + 1 \right) \log{(6)} = \left( 2x - 3 \right) \log{(2)} \end{align*}.

4. ## Re: Exponents and Logarithms-- 3 questions.

Originally Posted by Gurp925
Hi there i dont know if i should make separate threads maybe some one could let me know if so but here are three question i wanted to check if i was right or wrong without further adieu

1) Graph the following function: $\displaystyle f(x)=-\log(x+2)-1$ when i usually see a log function graph question i change to exponential form and when i do i get $\displaystyle 10^1=-(x+2)$ should it be $\displaystyle y+1=-(x+2)$ ?
No "logarith" is not linear and neither is an exponent.

2)Solve the following question leave in exact form: $\displaystyle (6^{3x+1})=(2^{2x-3})$ which i turn into $\displaystyle 3x+1\log(6)=2x-3\log(2)$ then i factor it out and bring x's together and solve from there i get $\displaystyle x= 1/24$
Did you try just putting 1/24 into the equation? Even without using a calculator it should be easy to see that (3/24)+ 1 is positive (so 6 to that power is greater than 1) while (2/24)- 1 is negative (so 2 to that power is less than 1).

5. ## Re: Exponents and Logarithms-- 3 questions.

Here is the graph