How do I solve this Exponential and Logarithmic Function?

1.Find the value of http://www.mathway.com/math_image.as...MB03?p=30?p=22 when http://www.mathway.com/math_image.as...MB03?p=34?p=22 = 4.

My attempt: http://www.mathway.com/math_image.as...MB03?p=30?p=22=http://www.mathway.com/math_image.as...MB03?p=34?p=22= 4

http://www.mathway.com/math_image.as...MB03?p=58?p=22=4

27x=1

x= 1/27.

When checked on a calculator it doesn't work. I don't understand what to do. Is there a way to solve this without using calculators?

2. Solve the equation http://www.mathway.com/math_image.as...013?p=138?p=22

I don't have an attempt since I can't understand what operation to use.

Whoops, wrong category. Can I request for a Moderator (if there is one in here) to relocate this to Algebra?

Now how do I mark this thread as [SOLVED]? There was a prefix box when I was just making this thread earlier.

Re: How do I solve this Exponential and Logarithmic Function?(Sorry, wrong category)

Quote:

Originally Posted by

**RPSGCC733**

$\displaystyle \displaystyle \begin{align*} 3^{9x} &= \left( 3^3 \right)^{3x} \\ &= 27^{3x} \\ &= \left( 27^x \right)^3 \\ &= 4^3 \\ &= 64 \end{align*}$

Re: How do I solve this Exponential and Logarithmic Function?(Sorry, wrong category)

Quote:

Originally Posted by

**RPSGCC733**

Is the second question trying to solve for $\displaystyle \displaystyle \begin{align*} x \end{align*}$ when $\displaystyle \displaystyle \begin{align*} \log_x{\left( \log_2{256} \right)} = 3 \end{align*}$?

Re: How do I solve this Exponential and Logarithmic Function?

Note that

$\displaystyle (3^{9})^x = (3^{3\cdot 3})^x = 3^{3\cdot 3 \cdot x} = \ldots = ?$

$\displaystyle \log_x(\log_2(256))=3$ where $\displaystyle \log_2(256)=8$ thus you have to solve

$\displaystyle \log_x(8)=3 \Leftrightarrow x^3 = 8 \Leftrightarrow \ldots $

Re: How do I solve this Exponential and Logarithmic Function?(Sorry, wrong category)

Quote:

Originally Posted by

**Prove It** $\displaystyle \displaystyle \begin{align*} 3^{9x} &= \left( 3^3 \right)^{3x} \\ &= 27^{3x} \\ &= \left( 27^x \right)^3 \\ &= 4^3 \\ &= 64 \end{align*}$

Ah. Thanks! Didn't think that 3^{9x} was 3^3 raised to the third power all along.

Yeah, it was trying to solve for x.

Re: How do I solve this Exponential and Logarithmic Function?

1.) A similar approach to the one already posted:

$\displaystyle 27^x=4$

$\displaystyle 3^{3x}=4$

$\displaystyle 3x=\log_3(4)$

$\displaystyle 9x=\log_3(64)$

$\displaystyle 3^{9x}=3^{\log_3(64)}=64$

Re: How do I solve this Exponential and Logarithmic Function?

Quote:

Originally Posted by

**Siron** Note that

$\displaystyle (3^{9})^x = (3^{3\cdot 3})^x = 3^{3\cdot 3 \cdot x} = \ldots = ?$

$\displaystyle \log_x(\log_2(256))=3$ where $\displaystyle \log_2(256)=8$ thus you have to solve

$\displaystyle \log_x(8)=3 \Leftrightarrow x^3 = 8 \Leftrightarrow \ldots $

Many thanks.

Isn't multiplication in Logarithm denoted with a + sign? That part confused me.

Nevermind, I just realized it wasn't multiplication. $\displaystyle \log_2(256)=8$ was a logarithm of $\displaystyle \log_x(y)=8$

Re: Exponential and Logarithmic Function?(Sorry, wrong category)

Hello, RPSGCC733!

Quote:

$\displaystyle \text{2. Solve: }\:\log_x\left[\log_2(256)\right] \:=\:3$

$\displaystyle \begin{array}{ccc}\text{Given:} & \log_x\left[\log_2(256)\right] \:=\:3 \\ \\ \text{Then:} & x^3 \:=\:\log_2(256) \\ \\ & x^3 \:=\:\log_2(2^8) \\ \\ & x^3 \:=\:8 \\ \\ & x \:=\:2 \end{array}$