For positive real numbers and with , the following equations are true: Which choice gives the value of ?

Solution available with step-by-step explanation

Super-Hard SAT Math Question 172 | Free SAT Prep | Aniko

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Question 172·200 Super-Hard SAT Math Questions·Advanced Math

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For positive real numbers $m$ and $r$ with $m\ne 1$ , the following equations are true:

m^{\frac{4}{5}}=r^{-\frac{3}{10}}

m^{6c-2}=r^2

Which choice gives the value of $c$ ? Аnikο.ai -⁠ ЅАТ‌ Preр

When you see two equations involving rational exponents and a parameter in an exponent, aim to express everything using a single base (here, rewrite $r$ as a power of $m$ ). Then substitute so both sides become $m^{\text{exponent}}$ . With $m>0$ and $m\ne 1$ , you can equate exponents directly, which is usually faster and cleaner than taking logarithms or expanding anything. © аnіko.ai

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Isolate one variable as a power

Use the first equation to rewrite $r$ as $m^{\text{(something)}}$ by raising both sides to an appropriate power. А-n-і-k-o.ai

Substitute into the second equation

Once you have $r=m^{k}$ , compute $r^2$ and replace $r^2$ in the equation $m^{6c-2}=r^2$ .

Use the fact that $m\ne 1$

If $m^{A}=m^{B}$ and $m>0$ with $m\ne 1$ , then $A=B$ . Apply that to the exponents after substitution.

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Pick a valid value for $m$

Set $m=2$ (any positive value other than 1 works).

Define $r$ from the first equation

Enter

m=2
r=(m^(4/5))^(-10/3)

This uses the idea that raising both sides of $m^{4/5}=r^{-3/10}$ to $-10/3$ isolates $r$ .

Create expressions for both sides of the second equation

Make a slider for c, then enter

L=m^(6c-2)
R=r^2

Adjust the slider to satisfy the equation

Adjust c until L and R are equal (or graph y=L and y=R and make them overlap). The corresponding slider value of c is the answer. аnіko.аi ⁠ЅAТ Quеѕti‌оn Вanк

Step-by-step Explanation

Rewrite $r$ in terms of $m$

Start with

m^{\frac{4}{5}}=r^{-\frac{3}{10}}.

Raise both sides to the power $-\frac{10}{3}$ so the exponent on $r$ becomes 1:

\left(m^{\frac{4}{5}}\right)^{-\frac{10}{3}}=\left(r^{-\frac{3}{10}}\right)^{-\frac{10}{3}}=r.

Simplify the exponent on $m$ :

\left(m^{\frac{4}{5}}\right)^{-\frac{10}{3}}=m^{\left(\frac{4}{5}\right)\left(-\frac{10}{3}\right)}=m^{-\frac{8}{3}}.

So $r=m^{-\frac{8}{3}}$ .

Compute $r^2$ as a power of $m$

r^2=\left(m^{-\frac{8}{3}}\right)^2=m^{-\frac{16}{3}}.

Substitute into the second equation and equate exponents

Substitute $r^2=m^{-\frac{16}{3}}$ into $m^{6c-2}=r^2$ :

m^{6c-2}=m^{-\frac{16}{3}}.

Because $m>0$ and $m\ne 1$ , equal powers of $m$ must have equal exponents, so

6c-2=-\frac{16}{3}.

Solve for $c$

Solve:

\begin{aligned} 6c-2&=-\frac{16}{3} \\ 6c&=-\frac{16}{3}+2=-\frac{16}{3}+\frac{6}{3}=-\frac{10}{3} \\ c&=\frac{-\frac{10}{3}}{6}=-\frac{10}{18}=-\frac{5}{9}. \end{aligned}

Therefore, the correct choice is $-\frac{5}{9}$ .

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Step-by-step Explanation

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