How do I solve this Exponential and Logarithmic Function?

**RPSGCC733** · October 14th 2012, 01:11 AM

1.Find the value of

when

= 4.

My attempt:

=

= 4

=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

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.

**Prove It** · October 14th 2012, 01:18 AM

Originally Posted by RPSGCC733

1.Find the value of

when

= 4.

My attempt:

=

= 4

=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

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?

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

**Prove It** · October 14th 2012, 01:20 AM

Originally Posted by RPSGCC733

1.Find the value of

when

= 4.

My attempt:

=

= 4

=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

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?

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

**Siron** · October 14th 2012, 01:22 AM

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

$\log_x(\log_2(256))=3$ where $\log_2(256)=8$ thus you have to solve
$\log_x(8)=3 \Leftrightarrow x^3 = 8 \Leftrightarrow \ldots$

**RPSGCC733** · October 14th 2012, 01:24 AM

Originally Posted by Prove It

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

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.

**MarkFL** · October 14th 2012, 01:27 AM

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

$27^x=4$

$3^{3x}=4$

$3x=\log_3(4)$

$9x=\log_3(64)$

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

**RPSGCC733** · October 14th 2012, 01:32 AM

Originally Posted by Siron

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

$\log_x(\log_2(256))=3$ where $\log_2(256)=8$ thus you have to solve
$\log_x(8)=3 \Leftrightarrow x^3 = 8 \Leftrightarrow \ldots$

Many thanks.

Isn't multiplication in Logarithm denoted with a + sign? That part confused me.
Nevermind, I just realized it wasn't multiplication. $\log_2(256)=8$ was a logarithm of $\log_x(y)=8$

**Soroban** · October 14th 2012, 02:06 PM

Hello, RPSGCC733!

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

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

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