Logarithms - GRE Subject Test: Math

Card 0 of 17

Question

Evaluate:

$3^{3x+3}=27^{2x+2}$

Answer

In order to evaluate the unknown variable, it is necessary to change the base. Looking at the right side of the equation, 27 is equivalent to three cubed.

Therefore, converting the right side of the equation to a base of 3 will allow setting both the left and right side of the exponential terms equal to each other.

$3^{3x+3}=27^{2x+2}$

$3^{3x+3}=3^{3(2x+2)}$

Log both sides to drop the exponents by log properties, and divide the log based 3 on both sides to cancel this term.

Solve for x.

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Question

Evaluate the following logarithm:

$log_{377}(563)$

Answer

The simplest way to evaluate a logarithm that doesn't have base 10 is with change of base formula:

$log_ba=\frac{log_ca}{log_cb}$

So we have

$log_{377}(563)$

$log_{377}563=\frac{log_{10}563}{log_{10}377}\approx1.068$

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Question

$Solve\ for\ x: 3^x=23$

Answer

In order to solve a logarithm, we must first rewrite it in log form:

$3^x=23\ written\ in\ log\ form\ is: log_{3}(23)=x$

To solve for x, we must use the Change of Base:

$log_{3}(23)=\frac{log(23)}{log(3)}$

$\frac{log(23)}{log(3)}=2.85405$

This means that:

$3^{2.85405}=23$

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Question

Solve:

Answer

Step 1: Re-write the log equation as an exponential equation. To do this, take the base of the log function and raise it to the number on the right side of the equal sign. This new exponent is equal to the number to the right of the log base.

Step 2: Re-write the right hand side as a power of 4..

Step 3: Re-write the equation

Step 4: We have the same base, so we can equal the exponents..

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Question

Solve for :

Answer

Use rules of logarithms...

Take the base of the log and raise it to the number on the right side of the equal sign (which becomes the exponent):

$\4^x=16^3 \4^x=(4^2)^3 \4^x=4^6 \x=6$

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Question

$Solve\ for\ x: log_{x}(49)=2$

Answer

In order to solve for x, we must first rewrite the log in exponential form.

Every log is written in the below general form:

$log_{base}(result)=exponent\ is\ rewritten\ as: base^e^x^p^o^n^e^n^t=result$

In this case we have:

$log_{x}(49)=2$

This becomes:

We can solve this by taking the square root of both sides:

$\sqrt{x^2}=\sqrt{49}$

$x=\pm7$

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Question

Evaluate: $\log_6 46656=x$

Answer

Step 1: Write the expression in exponential form...

Given: $\log_a b=x\implies a^x=b$

$\log_6 46656=x\implies 6^x=46656$

Step 2: Convert the right hand side into a power of 6..

Step 3: Re-write the equations...

Since the bases are equal, taking log of both sides will cancel them.

$\log_6 6^x=\log_6 6^6$

$\implies x=6$

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Question

$Solve\ for\ x:\ log_{2}(8)=x$

Answer

$In\ order\ to\ solve\ for\ x\ we\ must\ know\ how\ to\ rewrite\ a\ log\ in\ exponent\ form:$

$We\ can\ use\ the\ general\ rule:\ log_{b}(a)=x\ when\ rewritten\ in\ exponential\ form\ we\ get: b^{x}=a$

$To\ Solve\ for\ x:\ log_{2}(8)=x;\ we\ rewrite\ as\ 2^{x}=8$

$We\ know\ that\ 2^{3}=8;\ so\ the\ value\ of\ x=3$

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Question

Rewrite the following expression as a single logarithm

$5log_35x+log_3\frac{1}{4x^3}-log_3x^2$

Answer

Recall a few properties of logarithms:

1.When adding logarithms of like base, we multiply the inside.

2.When subtracting logarithms of like base, we divide the inside.

3. When multiplying a logarithm by a number, we can raise the inside to that power.

So we begin with this:

$5log_35x+log_3\frac{1}{4x^3}-log_3x^2$

I would start with 3 to simplify the first log.

$log_3(5x)^5+log_3\frac{1}{4x^3}-log_3x^2$

$log_33125x^5+log_3\frac{1}{4x^3}-log_3x^2$

Next, use rule 1 on the first two logs.

$log_3\left(3125x^5\cdot\frac{1}{4x^3}\right)-log_3x^2$

$log_3\left(\frac{3125x^2}{4}\right)-log_3x^2$

Then, use rule 2 to combine these two.

$log_3\left(\frac{3125x^2}{4x^2}\right)=log_3\left(\frac{3125}{4}\right)\approx 6.06$

So our answer is 6.06.

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Question

$Rewrite\ as\ one\ log: [log(x)+log(z)]-log(y)$

Answer

When combining logarithms into one log, we must remember that addition and multiplication are linked and subtraction and division are linked.

$log\left (\frac{a}{b} \right )=log(a)+log(b)$

In this case we have multiplication and division - so we assume anything that is negative, must be placed in the bottom of the fraction.

$log(xz)-log(y)=log\left ( \frac{xz}{y} \right )$

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Question

$Rewrite\ in\ log\ form: 12^(^x^+^2^)=y$

Answer

When rewriting an exponential function as a log, we must follow the model below:

$log_{base}(result)=exponent$

A log is used to find an exponent. The above corresponds to the exponential form below:

$12^(^x^+^2^)=y\ rewritten\ in\ log\ form\ becomes:$

$log_{12}(y)=x+2$

$The\ base\ is\ 12,\ the\ exponent\ is\ (x+2)\ and\ the\ result\ is\ y.$

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Question

$Rewrite\ in\ exponential\ form: log_{3}(22)=y$

Answer

In order to rewrite a log, we must remember the pattern that they follow below:

$log_{base}(result)=exponent$

In this question we have:

$log_{3}(22)=y$

$The\ base\ is\ 3,\ the\ exponent\ is\ y\ and\ the\ result\ is\ 22.$

$We\ now\ rewrite\ this\ as:\ 3^y=22$

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Question

Express $4\log_4 (x^2)-\log_4(y^{3})+\frac {1}{5}\log_4(z)$ as a single logarithm.

Answer

Step 1: Recall all logarithm rules:

$\log_a b+\log_a c=\log_a (bc)$

$\log_a b-\log_a c=log_a \bigg(\frac{b}{c}\bigg)$

$c\log_a b=\log_a (b^c)$

$\large a^{\frac {1}{b}}=\sqrt[b]{a}$

Step 2: Rewrite the first term in the expression..

$4\log_4 x^2=\log_4 (x^2)^4=log_4 (x^8)$

Step 3: Re-write the third term in the expression..

$\frac {1}{5}\log_4 (z)=\log_4 (z)^{\frac {1}{5}}=log_4 (\sqrt[5] {z})$

Step 4: Add up the positive terms...

$\log_4(x^8)+\log_4 (\sqrt[5]{z})=\log_4 (x^8\sqrt[5]{z})$

Step 5: Subtract the answer the other term from the answer in Step 4.

$\log_4 (x^8\sqrt[5]{z})-\log_4(y^3)=\log_4 \bigg(\frac {x^8\sqrt[5]{z}}{y^3}\bigg)$

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Question

$Expand: log\left ( \frac{xy}{z} \right )$

Answer

In order to expand this log, we must remember the log rules.

$Similar\ to\ the\ exponent\ rules\ when\ variables\ are\ multiplied\ within\ a\ log:\ the\ variables\ are\ added: log(xy)=log(x)+log(y)$

$When\ variables\ are\ divided\ within\ a\ log\ they\ are\ subtracted:\ log\left ( \frac{x}{y} \right )=log(x)-log(y)$

$For\ this\ problem\ we\ combine\ both\ rules:$

$Expand: log\left ( \frac{xy}{z} \right )=log(x)+log(y)-log(z)$

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Question

$Rewrite\ as\ one\ log:\ log(a)-(\left log(b)+log(c) \right)$

Answer

$When\ condensing\ logs\ we\ must\ remember\ the\ log\ properties:$

$log(x)+log(y)=log(xy): When\ adding;\ we\ multiply.$

$log(x)-log(y)=log\left (\frac{x}{y} \right ): When\ subtracting;\ we\ divide.$

$GIVEN: Rewrite\ as\ one\ log:\ log(a)-(\left log(b)+log(c) \right)\$

$This\ is\ rewritten\ as:\ log \left (\frac{a}{bc} \right )\ because\ the\ negative\ will\ be\ distributed\ to\ log(b)\ and\ log(c)$

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Question

$Solve\ for\ x: log_{3}(x)=2$

Answer

$We\ can\ use\ the\ general\ rule:\ log_{b}(a)=x\ when\ rewritten\ in\ exponential\ form\ we\ get: b^{x}=a$

$In\ this\ case\ we\ have\ log_{3}(x)=2$

$This\ can\ be\ rewritten\ in\ exponential\ form\ as: 3^2=x$

$We\ know\ that\ 3^2=9;\ so\ x=9$

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Question

Rewrite the following expression as a single logarithm

$5log_35x+log_3\frac{1}{4x^3}-log_3x^2$

Answer

Recall a few properties of logarithms:

1.When adding logarithms of like base, we multiply the inside.

2.When subtracting logarithms of like base, we divide the inside.

3. When multiplying a logarithm by a number, we can raise the inside to that power.

So we begin with this:

$5log_35x+log_3\frac{1}{4x^3}-log_3x^2$

I would start with 3 to simplify the first log.

$log_3(5x)^5+log_3\frac{1}{4x^3}-log_3x^2$

$log_33125x^5+log_3\frac{1}{4x^3}-log_3x^2$

Next, use rule 1 on the first two logs.

$log_3\left(3125x^5\cdot\frac{1}{4x^3}\right)-log_3x^2$

$log_3\left(\frac{3125x^2}{4}\right)-log_3x^2$

Then, use rule 2 to combine these two.

$log_3\left(\frac{3125x^2}{4x^2}\right)=log_3\left(\frac{3125}{4}\right)\approx 6.06$

So our answer is 6.06.

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