If #Log_y6=x#, #Log_y3.5=z# and #6a=3.5# then #a=??#

Question

If #Log_y6=x#, #Log_y3.5=z# and #6a=3.5# then #a=??#

#(a)# #xz#
#(b)# #z/x#
#(c)# #z+x#
#(d)# #z-x#

I searched on internet about correct answer and it was #(b)# #z/x#, but I don't know why #(b)# is correct. I think that all the answers listed are incorrect.

Eli Assouline · Accepted Answer

I got #a=y^z/y^x=y^(z-x)#

#6a=3.5=>a=3.5/6#

We're told that

#log_y6=x; log_y3.5=z#

We can rearrange these equations to put them into exponential form:

#y^x=6; y^z=3.5#

And substitute in the values:

#a=y^z/y^x=y^(z-x)#

Savannah Anderson · Answer

undefined Given that $ \log_y{6} = x $ and $ \log_y{3.5} = z $, and that $ 6a = 3.5 $, to find the value of $ a $, we first express $ 6 $ and $ 3.5 $ in terms of $ y $ using the given logarithmic expressions. Then, we solve for $ a $.

Since $ 6 = y^x $ and $ 3.5 = y^z $, we have:

$ 6a = 3.5 $

Substituting the expressions for $ 6 $ and $ 3.5 $, we get:

$ y^x \times a = y^z $

To solve for $ a $, we divide both sides by $ y^x $:

$ a = \frac{y^z}{y^x} $

Using the properties of logarithms, we can rewrite this as:

$ a = y^{z-x} $

Thus, the value of $ a $ in terms of $ y $, $ x $, and $ z $ is $ y^{z-x} $.

HIX Tutor · Answer

undefined Great question! Let's break it down step by step.

1. **Understand the logs**: You have two equations:
   - $ \log_y 6 = x $
   - $ \log_y 3.5 = z $

What do you think $ y^x $ equals?

2. **Rewrite the logs**:
   - From $ \log_y 6 = x $, we can say $ y^x = 6 $.
   - From $ \log_y 3.5 = z $, we can say $ y^z = 3.5 $.

3. **Look at the equation**: You also have $ 6a = 3.5 $.

4. **Solve for $ a $**:
   - You want to find $ a $. Rewrite the equation: 
     $$
     a = \frac{3.5}{6}
     $$

5. **Divide**: Now we can divide $ 3.5 $ by $ 6 $:
   - What is $ 3.5 \div 6 $? 
   - $ 3.5 $ is the same as $ \frac{35}{10} $.
   - Now let's simplify: 
     $$
     \frac{3.5}{6} = \frac{35/10}{6} = \frac{35}{60}
     $$

6. **Reduce the fraction**: 
   - Both $ 35 $ and $ 60 $ can be divided by $ 5 $:
     - $ 35 \div 5 = 7 $
     - $ 60 \div 5 = 12 $
   - So, $ \frac{35}{60} = \frac{7}{12} $.

7. **Final answer**: 
   - Therefore, $ a = \frac{7}{12} $.

Does this make sense? Would you like to go over any part again?

If #Log_y6=x#, #Log_y3.5=z# and #6a=3.5# then #a=??#

If #Log_y6=x#, #Log_y3.5=z# and #6a=3.5# then #a=??# #(a)# #xz# #(b)# #z/x# #(c)# #z+x# #(d)# #z-x# I searched on internet about correct answer and it was #(b)# #z/x#, but I don't know why #(b)# is correct. I think that all the answers listed are incorrect.

If #Log_y6=x#, #Log_y3.5=z# and #6a=3.5# then #a=??#

#(a)# #xz#
#(b)# #z/x#
#(c)# #z+x#
#(d)# #z-x#

I searched on internet about correct answer and it was #(b)# #z/x#, but I don't know why #(b)# is correct. I think that all the answers listed are incorrect.